Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling...

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Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000 Haifa, ISRAEL September 3, 2018 Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutat

Transcript of Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling...

Page 1: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Touchard Polynomials, Stirling Numbers andRandom Permutations

Ross Pinsky

Department of Mathematics, Technion32000 Haifa, ISRAEL

September 3, 2018

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 2: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 3: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 4: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 5: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 6: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 7: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Notation:

[n] = {1, 2, · · · , n}

Sn = permutations of [n]

|Sn| = n!

S6 3 σ1 =

(1 2 3 4 5 62 4 6 1 5 3

)= (124)(36)(5),

S6 3 σ2 =

(1 2 3 4 5 63 4 6 5 1 2

)= (136245) = (624513)

σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.

There are (n − 1)! different n-cycles in Sn.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 8: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)

(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 9: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 10: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n

Recurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 11: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 12: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|.

Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 13: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Rising Factorials

x (n) := x(x +1) · · · (x +n−1)(≡

n∑k=1

ankxk), x ∈ R, n ≥ 1. (1)

Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation

|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;

|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)

It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore

x (n) =n∑

k=1

|s(n, k)|xk . (3)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 14: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 15: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) =

(−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 16: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)

= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 17: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).

Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 18: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k

=

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 19: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 20: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind.

From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 21: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Falling Factorials

(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)

Substituting −x for x in (4), we have

(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have

(x)n = (−1)n(−x)(n)from (3)

= (−1)nn∑

k=1

|s(n, k)|(−x)k =

n∑k=1

(−1)n−k |s(n, k)|xk .(5)

Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have

(x)n =n∑

k=1

s(n, k)xk . (6)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 22: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.

Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 23: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:

{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 24: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 25: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.

Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 26: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 27: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.

How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 28: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?

The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 29: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 30: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:

For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 31: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)

So the answer by this indirect count is∑n

k=1 ck .Thus, xn =

∑nk=1 ck . To complete the proof we now show that

ck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

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Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 33: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck .

To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 34: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Stirling Numbers of the Second Kind

S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.

xn =n∑

k=1

S(n, k)(x)k , (7)

Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.

An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is

∑nk=1 ck .

Thus, xn =∑n

k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 35: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 36: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}.

Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 37: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 38: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 39: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 40: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 41: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k!

= S(n, k) x!(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 42: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)!

= S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 43: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :

For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then

f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.

Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};

2. There are(xk

)was to choose the {x1, · · · , xk};

3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.

Thus ck = S(n, k)×(xk

)× k! = S(n, k) x!

(x−k)! = S(n, k)(x)k . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 44: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 45: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 46: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 47: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).

Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 48: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).

Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 49: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 50: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1

(x)n =∑n

k=1 s(n, k)xk , n ≥ 1

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 51: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof.

Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 52: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R.

.

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 53: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 54: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 55: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

xn =∑n

k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =

∑nk=1 s(n, k)xk , n ≥ 1 (**)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform

∑nk=1 ckx

k , n ≥ 1, ck ∈ R. .

Of course,B1 := {x , x2, x3, · · · } is a basis for V .

By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .

By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 56: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.

Example: n = 8,m = 4:0 =

∑∞j=1 S(8, j)s(j , 4) =

S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 57: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets

|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles

s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)

Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.

Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id

That is,∑∞

j=1 S(n, j)s(j ,m) =

{1, if n = m;

0, if n 6= m.Example: n = 8,m = 4:0 =

∑∞j=1 S(8, j)s(j , 4) =

S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 58: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Recalling the Poisson distribution

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 0, 1, 2, · · · .

nth moment: µn;λ := EX n

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 59: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Recalling the Poisson distribution

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 0, 1, 2, · · · .

nth moment: µn;λ := EX n

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 60: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k

=

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 61: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k

(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 62: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 63: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 64: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)!

= λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 65: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)!

= λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 66: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l!

=

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 67: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ.

Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 68: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∞∑j=0

(e−λ

λj

j!

)jn

(7)=∞∑j=0

(e−λ

λj

j!

) n∑k=1

S(n, k)(j)k =

=∞∑j=0

(e−λ

λj

j!

)min(j ,n)∑k=1

S(n, k)(j)k(because (j)k = 0, k > j)

= e−λn∑

k=1

S(n, k)∞∑j=k

λj

j!(j)k .

But∑∞

j=kλj

j! (j)k =∑∞

j=kλj

(j−k)! = λk∑∞

j=kλj−k

(j−k)! = λk∑∞

l=0λl

l! =

λkeλ. Thus,

µn;λ =n∑

k=1

S(n, k)λk .

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 69: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 70: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 71: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)

(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 72: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 73: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 74: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k)

:= Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 75: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =

∑∞j=0

(e−λ λj

j!

)jn

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)(Compare Tn with (6).)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;1 = Tn(1) :=

n∑k=1

S(n, k) := Bn,

where Bn is the nth Bell number:Bn = the total number of partitions of [n].

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

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Let X be a random variable with the distribution Poiss(λ), λ > 0:

P(X = j) = e−λ λj

j! , j = 1, 2, · · · .nth moment: µn;λ := EX n

µn;λ =∑n

k=1 S(n, k)λk

Touchard Polynomial: Tn(x) :=∑n

k=1 S(n, k)xk

µn;λ = Tn(λ)

∞∑j=0

(e−λ

λj

j!

)jn = EX n = µn;λ = Tn(λ) :=

n∑k=1

S(n, k)λk

When λ = 1:

1

e

∞∑j=0

jn

j!= EX n = µn;λ = Tn(1) :=

n∑k=1

S(n, k) := Bn, (8)

where Bn is the nth Bell number:Bn = the total number of partitions of [n].(8) is known as Dobinski’s formula.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 77: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 78: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 79: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 80: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 81: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 82: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)|

(3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 83: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 84: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 85: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Random Permutations

Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn

We now define a family of non-uniform probability measures on Sn:

Let kn(σ) = the number of cycles in σ.

For θ > 0, define the following probability measure on Sn:

P(θ)n (σ) = θkn(σ)

Nn,θ, where Nn,θ =

∑σ∈Sn θ

kn(σ) is the normalizingconstant.

We have Nn,θ =∑

σ∈Sn θkn(σ) =

∑nk=1 θ

k |s(n, k)| (3)= θ(n).

Thus, P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 86: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure

Let C(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 87: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 88: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 89: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞.

That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 90: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 91: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

P(θ)n (σ) = θkn(σ)

θ(n), σ ∈ Sn.

θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C

(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.

Then for each θ > 0, we can think of {C (n)j }nj=1 as random

variables on the probability space (Sn,P(θ)n ).

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θj( θj)m

m! , for m = 0, 1, · · · .

Remark. C(n)1 (σ) is the number of fixed points of σ:

C(n)1 = |{k ∈ [n] : σk = k}|. So under P

(θ)n , the distribution of

the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 92: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θm

( θj)m

m! , for m = 0, 1, · · · .

Proof for θ = 1 (the same method works with virtually no extrawork for all θ).

Method of moments: It is enough to show that for all m, the

mth moment of C(n)j converges to the mth moment of the

distribution Poiss(1j ).

That is, it is enough to show that

limn→∞

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 93: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θm

( θj)m

m! , for m = 0, 1, · · · .

Proof for θ = 1 (the same method works with virtually no extrawork for all θ).

Method of moments: It is enough to show that for all m, the

mth moment of C(n)j converges to the mth moment of the

distribution Poiss(1j ).

That is, it is enough to show that

limn→∞

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 94: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Theorem. For any j, the distribution of C(n)j under P

(θ)n converges

weakly to the distribution Poiss( θj ) as n→∞. That is,

limn→∞ P(θ)n (C

(n)j = m) = e−

θm

( θj)m

m! , for m = 0, 1, · · · .

Proof for θ = 1 (the same method works with virtually no extrawork for all θ).

Method of moments: It is enough to show that for all m, the

mth moment of C(n)j converges to the mth moment of the

distribution Poiss(1j ).

That is, it is enough to show that

limn→∞

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 95: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 96: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 97: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 98: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ)

and(C

(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 99: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 100: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

Proof that limn→∞ E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1j

).

We will show that

E(1)n

(C

(n)j

)m= µm; 1

j= Tm

(1

j

), for n ≥ mj .

For D ⊂ [n] with |D| = j , define

1D(σ) =

{1, if σ has a cycle consisting of the elements of D;

0, otherwise.

Then C(n)j (σ) =

∑D⊂[n]:|D|=j 1D(σ) and(

C(n)j (σ)

)m=( ∑

D1⊂[n]:|D1|=j

1D1(σ))· · ·( ∑

Dm⊂[n]:|Dm|=j

1Dm(σ))

=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

m∏l=1

1Dl(σ).

SoE(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 101: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 102: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0,

(that is, there exists some σ ∈ Dn such that∏ml=1 1Dl

(σ) 6= 0),if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 103: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 104: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 105: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 106: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai,

and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 107: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 108: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=

∑(D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

( m∏l=1

1Dl= 1).

Now∏m

l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m

l=1 1Dl(σ) 6= 0),

if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.

That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.

In this case,∏m

l=1 1Dl=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 109: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .

The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 110: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 111: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k).

So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 112: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k)

=

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 113: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k)

= Tm(1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 114: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j)

= µm; 1j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations

Page 115: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000

E(1)n

(C

(n)j

)m=∑(

D1,D2,··· ,Dm

):|Dl |=···=|Dm|=j

P(1)n

(∏ml=1 1Dl

= 1).

Now∏m

l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint

sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl

=∏k

i=1 1Ai, and

P(1)n

(∏ml=1 1Dl

= 1)

= P(1)n

(∏ki=1 1Ai

= 1)

=

((j−1)!

)k(n−kj)!

n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj

)(n−jj

)· · ·(n−(m−1)j

j

)= n!

(j!)k (n−jk)! .

Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So

E(1)n

(C

(n)j

)m=

m∑k=1

((j − 1)!

)k(n − kj)!

n!× n!

(j!)k(n − jk)!× S(m, k) =

m∑k=1

(1

j

)kS(m, k) = Tm(

1

j) = µm; 1

j.

Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations