MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 Facts Recall That, the Stirling Number S(K, N) of the Second Kind Is Defined As

MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2

KANNAPPAN SAMPATH

Facts Recall that, the Stirling number S(k, n) of the second kind is deﬁned as the number of partitions of a [k] into n non-empty blocks. We recall the following result from class:

Lemma 0.1. The following numbers are equal: (1) The number of surjective maps from [k] to [n]. (2) ∑S(k, n)n! where( ) S(k, n) denotes the Stirling number of the second kind. n − i n − k (3) i=0( 1) i (n i) . Proof. (1) = (2): Given a partition of [k] into n blocks, one may deﬁne a surjective map by mapping distinct parts to distinct numbers between 1 and n. There are n! such surjections. (1) = (3): (Principle of Inclusion and Exclusion.) Let Ai denote the set of maps f :[k] → [n] such that i∈ / Im(f). Then, it follows that the number of surjective → maps [k] [n] is given by

∪k [k] \ [n] Ai . i=1 Applying the principle of inclusion-exclusion, we conclude:

∪k ∑ [k] \ − |I| [n] Ai = ( 1) AI i=1 I⊆[k] ( ) ∑n n = (−1)i (n − i)k. i i=0 This completes the proof. □

Problems

Problem 1. Let S(n, k) be the number of ways of partitioning an n(-set) into k n−1 − − n blocks. Prove that S(n, 1) = 1, S(n, 2) = 2 1 and S(n, n 1) = 2 , Find a formula for S(n, n − 2).

Solution. Proof that S(n, 1) = 1: There is exactly one partition of an n-set which consists of just one block, viz, [n]. So, S(n, 1) = 1. 1 2 KANNAPPAN SAMPATH

Proof that S(n, 2) = 2n−1 − 1: Let Σ(n, 2) denote the set of all partitions of [n] into 2 parts (so⨿ that |Σ(n, 2)| = S(n, 2)). Note that any element of Σ(n, 2) is of the form A Ac for a subset A of [n] such that A ≠ ∅ ≠ Ac. Consider the map: ϕ : 2[n] \{∅, [n]} → Σ(n, k) ⨿ A 7→ A Ac Then, this map is clearly surjective and is 2-to-1, indeed ϕ(A) = ϕ(Ac). Thus, 2|Σ(n, k)| = |2[n] \{∅, [n]}| = 2n − 2 − whence we have S((n,)2) = 2n 1 − 1 as required. − n − Proof that S(n, n 1) = 2 : Let Σ(n, n 1) denote the set of all partitions of [n] into n−1 parts (so that |Σ(n, n−⨿1)| =⨿S(n,⨿ n−1)). Note that any element of Σ(n, n − 1) is of the form A1 A2 ··· An−1 where Ai is non-empty for every i. For a ﬁxed partition of the above form, let ai = |Ai|. Then, we have that ai ⩾ 1 and n∑−1 ai = n. i=1

If there are atleast two indices, say i, j such that ai, aj ⩾ 2, then, we have that n∑−1 n = ai ⩾ 2 + 2 + (n − 1) − 2 = n + 1 i=1

a contradiction. On the other hand, if ai = 1 for all i, then, we get the absurd equation n = n − 1. Thus, there is precisely one index i such that ai = 2 and aj = 1 for all j ≠ i. We may let i = n − 1 without loss of generality. (This is an elaboration of the Pigeonhole principle.) The above consideration shows that an element of Σ(n, n − 1) is deter- mined by the 2-set An−1. Conversely, every 2-subset T of [n] determines a partition of [n] into n − 1 blocks: ⨿ ⨿ [n] = {a} T. a/∈T ( ) − n This shows that S(n, n 1) = 2 (the number of 2-subsets of [n]). Formula for S(n, n − 2): In this part, we argue as in the previous part, but some- what more systematically. Let kc denote the number of blocks with car- dinality c in a given partition of [n] into n − 2 blocks. Then, we have the equation: ∑ ∑ ∑ (n − 2) − kc + ckc = n =⇒ (c − 1)kc = 2. c⩾2 c⩾2 c⩾2

This equation shows that if kc > 0, then c ⩽ 3. Further it follows that k2 = 2, k3 = 0 or k2 = 0, k3 = 1. The number of partitions of [n] into n(−)(2 blocks) consisting of two blocks 1 n n−2 of size 2 (and other blocks singleton) is 2 2 2 . The number of partitions of [n] into n −( )2 blocks consisting of one block of size 3 (and other blocks n singleton) is 3 . MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 3

Based on the above analysis, it follows that ( ) ( )( ) n 1 n n − 2 n(n − 1)(n − 2)(3n − 5) S(n, n − 2) = + = . 3 2 2 2 24 This completes this problem. □

Remark 0.2. Another way to approach the above problem would be through recurrence established for Stirling numbers S(n, k) of the second kind. Recall from class that S(n, k) = S(n − 1, k − 1) + kS(n − 1, k)

Set an = S(n, n − 2); we know that a3 = S(3, 1) = 1. On taking k = n − 2 in the above recurrence relation, we have ( ) n − 1 a = a − + (n − 2) n n 1 2 using the previous calculation. Rearranging the above equality and summing from n = 4 to k gives us: ∑k ak − 1 = (an − an−1) n=4 ∑k (n − 1)(n − 2)2 = 2 n=4 Making a change of variable in the above sum: − k∑2 n2(n + 1) = 2 n=2 − 1 k∑2 = n3 + n2 2 n=2 Using the known formulae: ( ) 1 (k − 2)2(k − 1)2 (k − 2)(k − 1)(2k − 3) = − 1 + − 1 2 4 6 ( ) (k − 1)(k − 2) (k − 2)(k − 1) 2k − 3 a = + k 4 2 3 k(k − 1)(k − 2)(3k − 5) = 24 This also proves the expression we computed for S(n, n − 2) above.

Problem 2. Let fn and gn be two sequences with exponential generating functions ∞ ∞ ∑ f tn ∑ g tn F (t) = n ,G(t) = n . n! n! n=0 n=0 Show that the following are equivalent: (a) ( ) ∑n n g = f ; n k k k=0 4 KANNAPPAN SAMPATH

(b) G(t) = F (t)et.

Proof. Since an exponential generating function uniquely determines the sequence and conversely, the equivalence is immediate in view of the following calculation: ( ) ∑n n g = f n k k k=0 ∞ ( ) ∑ tn ∑n n ⇐⇒ G(t) = f n! k k n=0 k=0 ∞ ∑ tn ∑n n! ⇐⇒ G(t) = f n! k!(n − k)! k n=0 k=0 Interchanging the order of summation: ∞ ∞ ∑ tk ∑ tn−k ⇐⇒ G(t) = f k! k (n − k)! k=0 n=k Making the change of variable n − k = u in the inner sum: ∞ ∞ ∑ tk ∑ tu ⇐⇒ G(t) = f k! k u! k=0 u=0 ⇐⇒ G(t) = F (t)et This ﬁnishes the proof. □

Problem 3. Let dn be the number of derangements of an n-element set (that is, the number of elements of Sn without any ﬁxed points). Show that n dn = ndn−1 + (−1)

Deduce that dn is odd if and only if n is even.

Proof. We have shown in class that dn satisﬁes the following recurrence:

dn = (n − 1)(dn−1 + dn−2), for n > 1 d0 = 1, d1 = 0.

Substracting ndn−1, the relation becomes:

dn − ndn−1 = −(dn−1 − (n − 1)dn−2)

Setting cn = dn − ndn−1, we see that cn = −cn−1 with c1 = −1. Thus, it follows n immediately that cn = (−1) which proves the relation required. n Now, the ﬁnal parity argument is easy: if n is even, then, dn ≡ (−1) = 1 mod 2, so dn is odd. On the other hand, if n is odd, then, dn is even since n − 1 is even and we have just shown that dn−1 is odd; using the fact that dn = ndn−1 − 1, we have that dn is even. This ﬁnishes the problem. □

Problem 4. The Bernoulli numbers bn are deﬁned by the recurrence b0 = 1 and ( ) ∑n n + 1 b = 0 k k k=0 MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 5 for n ⩾ 1. Prove that ∞ ∑ b t f(t) := n tn = . n! et − 1 n=0 Proof. From the given recurrence, we have

− ( ) 1 n∑1 n + 1 b = − b n n + 1 k k k=0 for n > 0. Now, we compute the function f(t) as follows: ∞ ∑ b tn f(t) = n n! n=0 ∞ − ( ) ∑ tn n∑1 n + 1 = 1 − b (n + 1)! k k n=1 k=0 ∞ − ∑ n∑1 1 = 1 − tn b k!(n + 1 − k)! k n=1 k=0 Put m = n − 1: ∞ ∑ ∑m 1 = 1 − tm+1 b k!(m + 2 − k)! k m=0 k=0 Interchanging the order of summation, we have: ∞ ∞ ∑ ∑ tm+1 = 1 − b k!(m + 2 − k)! k k=0 m=k Put v = m − k + 2 to get: ∞ ∞ ∑ ∑ tv+k−1 = 1 − b k!v! k k=0 v=2 ∞ ∞ ∑ tk−1b ∑ tv = 1 − k k! v! k=0 v=2 f(t)(et − t − 1) f(t) = 1 − t t f(t) = . et − 1 This ﬁnishes the proof. □

Problem 5. In the previous question, show that t f(t) + 2 is an even function of t (that is, it is invariant under the map t 7→ −t). Deduce that bn = 0 for all odd n ⩾ 3. 6 KANNAPPAN SAMPATH

Proof. To prove the ﬁrst part, we simply compute: −t −t t f(−t) + = − 2 e−t − 1 2 2tet − t(et − 1) = 2(et − 1) tet + t = 2(et − 1) ( ) t et + 1 = 2 et − 1 ( ) t et − 1 + 2 = 2 et − 1 ( ) t 2 = 1 + 2 et − 1 t t = + 2 et − 1 t = f(t) + 2 Now, plugging our power series for f(t), we get: ∞ ∞ ∑ (−1)nb tn −t ∑ b tn t n + = n + n! 2 n! 2 n=0 n=0 n Comparing the coeﬃcient of t on both sides implies that b1 = −1/2 and that bn = 0 for all odd n ⩾ 3. □

Problem 6. Let s(n, k) and S(n, k) denote the Stirling numbers of the ﬁrst and second kind respectively. Show that S(n, k) ⩽ |s(n, k)|. Deduce that the n-th Bell number satisﬁes Bn ⩽ n!.

Proof. By deﬁnition, we have the following descriptions: S(n, k): the number of (unordered) partitions of [n] into k blocks. |s(n, k)|: the number of elements in Sn that can be written as a product of k disjoint cycles. Bn: the number of (unordered) partitions of [n]. Based on these descriptions, it is easy to deduce the claim about Bell numbers: indeed, we have ∑n Bn = S(n, k) k=1 ∑n ⩽ |s(n, k)| = n!. k=1 Thus, to ﬁnish the problem, we must show that S(n, k) ⩽ |s(n, k)|. We show that there is an injection from the set Σ(n, k) of partitions of [n] into k blocks into the set of elements in Sn that⨿ can be⨿ written as a product of k disjoint cycles. Indeed, given an element σ = σ[1] ··· σ[k] ∈ Σ(n, k), we may construct a permutation MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 7

σ˙ as follows:

σ˙ = (σ11 . . . σ1r1 ) ... (σk1 . . . σkrk ) ··· { } where σj1 < < σjrj and σ[j] = σj1, . . . , σjrj . This map is clearly an injection and this shows that S(n, k) ⩽ |s(n, k)|. □

Problem 7. With Bn denoting the n-th Bell number (that is, the number of partitions of an n-element set), show that   − ∑n kn n∑k (−1)j B =   . n k! j! k=1 j=0

Proof. We begin by noting ∑n Bn = S(n, k) k=1 where S(n, k) is the Stirling number of the second kind. Thus (for n > 0):

− ( ) ∑n 1 k∑1 k = (−1)i (k − i)n. k! i k=1 i=0 − ∑n k∑1 1 = (−1)i (k − i)n. i!(k − i)! k=1 i=0 Substituting j = k − i, it follows that:

∑n ∑k 1 = (−1)k−j jn (k − j)!j! k=1 j=1 Interchanging the order of summation, it follows that: ∑n ∑n 1 = (−1)k−j jn (k − j)!j! j=1 k=j ∑n jn ∑n 1 = (−1)k−j j! (k − j)! j=1 k=j Making a change of variable in the above sum, it follows that:

n−j ∑n jn ∑ 1 = (−1)k j! k! j=1 k=1 This ﬁnishes the proof. □

Problem 8. A permutation σ of the symmetric group Sn is called an involution if 2 σ = 1. Let s(n) denote the number of involutions of Sn. Deﬁning s(0) = 1, prove that ∞ ( ) ∑ s(n)tn t2 = exp t + . n! 2 n=0 8 KANNAPPAN SAMPATH

Proof. We ﬁrst establish a recurrence relation for s(n). Let σ ∈ Sn such that 2 σ = 1. If σ(n) = n, then, σ is an involution in Sn−1. Otherwise, (nσ(n))σ is an involution on n − 2 symbols. This gives us the following recurrence: s(n) = s(n − 1) + (n − 1)s(n − 2). ∑ ∞ s(n)tn Let S(t) = n=0 n! . We now do the following: ∞ ∑ s(n)tn (1 + t)S(t) = (1 + t) n! n=0 ∞ ∞ ∑ s(n)tn ∑ s(n)tn+1 = + n! n! n=0 n=0 ∞ ∞ ∑ s(n)tn ∑ s(n − 1)tn = + n! (n − 1)! n=0 n=1 ∞ ∞ ∑ s(n)tn ∑ ns(n − 1)tn = + n! n! n=0 n=1 ∞ ∑ s(n) + ns(n − 1) = 1 + tn n! n=1 ∞ ∑ s(n + 1) = 1 + tn n! n=1 ∞ ∑ ns(n)tn−1 = 1 + n! n=2 ∞ ∑ ns(n)tn−1 = n! n=1 = S′(t)

This proves that S(t) = exp(t + t2/2) since S(0) = 1. □

Problem 9. Prove that ( )( ) ( ) ∑n n m + n − i m (−1)i = i k − i k i=0 if k ⩽ m and 0 otherwise. [Hint: if you have m red cards and n blue cards, count the number of k-element subsets consisting solely of red cards.]

Proof. Continuing the notation in Hint, let A denote the set of all k-subsets of m + n cards. Index the blue cards from 1 to n. For each i ∈ [n], let Ai denote the set of k-subsets of m + n cards which consists of the ith blue card. The number of k-subsets of m + n cards which consists only of red cards is then given by: {( ) ∪n m ⩽ k , if k m A \ Ai = . 0, otherwise i=1 MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 9

Applying the Principle of Inclusion-Exclusion, we see:

∪m ∑ \ − |I| (1) A Ai = ( 1) AI i=1 I⊆[n] where AI = ∩i∈I Ai (with the obvious convention that A∅ = A). Clearly, AI is just the set of k-subsets of m + n cards which contains( all) the blue cards whose indices | | m+n−i are in I. If I = i, then, this number is just k−i . Notice that this number is independent of the subset I (depends only on |I|). Now, continuing from (1):

∪m ∑ \ − |I| A Ai = ( 1) AI i=1 I⊆[n] ( )( ) ∑n n m + n − i = (−1)i . i k − i i=0 This completes the proof. □

Problem 10. Let An denote the number of maps f :[n] → [n] with the property that if i lies in the image of f, then, so does [i]. Prove that ( ) ∑n n A = A − n k n k k=1 Deduce that ∞ ∑ kn A = n 2k+1 k=0

In this problem, we have (by convention, for the scholars of the empty set) that A0 = 1.

Proof. We reinterpret the number An. Firstly, a function f :[n] → [n] that satisﬁes the hypothesis of the problem must surject onto [k] for some 1 ⩽ k ⩽ n and conversely; now, a surjective function f :[n] → [k] determines an ordered partition of [n] into k blocks and conversely, given a partition ⨿ ⨿ ⨿ P : S1 S2 ··· Sk = [n] of [n] into k blocks, we may construct fP :[n] → [k] by fP (j) = ℓ if j ∈ Sℓ. Thus, An is the number of ordered partition of [n] into non-empty blocks (for n > 0); let An denote the set of ordered partitions of [n]. Given an ordered partition P of [n], we write P(1) for the ﬁrst block of the given partition. Then, we have the following count (assuming that n > 0): ∑ An = 1 P∈A ∑n ∑ = 1 ∅≠ S⊆[n] P[1]=S 10 KANNAPPAN SAMPATH

The inner sum counts the number of ordered partitions with first block S which is the same as the number of ordered partitions of [n − |S|]: ∑ = An−|S| ∅≠ S⊆[n] ( ) ∑n n = A − k n k k=1 This finishes the proof of the first equality. Now to prove the second equality, we proceed by induction on n. For n = 0, the result holds. Indeed, the sum defining A0 is a geometric series with initial term 1/2 and common ratio 1/2 whence its sum is 1 (by convention, as always, 00 = 1). Since the recursion holds for all n ⩾ 1, we plug this explicit formula into the recursion and get that: ( ) ∑n n A = A − n k n k k=1 ( ) ∞ ∑n n ∑ ℓn−k = k 2ℓ+1 k=1 ℓ=0 Interchanging the order of summation: ∞ ( ) ∑ 1 ∑n n = ℓn−k 2ℓ+1 k ℓ=0 k=1 We keep the term ℓ = 0 out (and implement our convention that 00 = 1): ∞ ( ) 1 ∑ ℓn ∑n n = + ℓ−k 2 2ℓ+1 k ℓ=1 k=1 ∞ (( ) ) 1 ∑ ℓn 1 n = + 1 + − 1 2 2ℓ+1 ℓ ℓ=1 ∞ ∞ 1 ∑ (ℓ + 1)n ∑ ℓn = + − 2 2ℓ+1 2ℓ+1 ℓ=1 ℓ=1 Making a change of variable in the first sum: ∞ ∞ 1 ∑ ℓn ∑ ℓn = + − 2 2ℓ 2ℓ+1 ℓ=2 ℓ=1 ∞ ∑ ℓn = 2ℓ+1 ℓ=1 This completes the problem. □