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) PS(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 i2 = Im(f). Then, it follows that the number of surjective ! maps [k] [n] is given by [k [k] n [n] Ai : i=1 Applying the principle of inclusion-exclusion, we conclude: [k X [k] n − jIj [n] Ai = ( 1) AI i=1 I⊆[k] ( ) Xn 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 jΣ(n; 2)j = 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 =6 ? =6 Ac. Consider the map: ϕ : 2[n] n f?; [n]g ! Σ(n; k) a A 7! A Ac Then, this map is clearly surjective and is 2-to-1, indeed ϕ(A) = ϕ(Ac). Thus, 2jΣ(n; k)j = j2[n] n f?; [n]gj = 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 jΣ(n; n−`1)j =`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 = jAij. Then, we have that ai > 1 and nX−1 ai = n: i=1 If there are atleast two indices, say i; j such that ai; aj > 2, then, we have that nX−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 =6 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: a a [n] = fag T: a=2T ( ) − 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: X X X (n − 2) − kc + ckc = n =) (c − 1)kc = 2: c>2 c>2 c>2 This equation shows that if kc > 0, then c 6 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 re- currence 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: Xk ak − 1 = (an − an−1) n=4 Xk (n − 1)(n − 2)2 = 2 n=4 Making a change of variable in the above sum: − kX2 n2(n + 1) = 2 n=2 − 1 kX2 = 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 1 1 X f tn X g tn F (t) = n ;G(t) = n : n! n! n=0 n=0 Show that the following are equivalent: (a) ( ) Xn 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: ( ) Xn n g = f n k k k=0 1 ( ) X tn Xn n () G(t) = f n! k k n=0 k=0 1 X tn Xn n! () G(t) = f n! k!(n − k)! k n=0 k=0 Interchanging the order of summation: 1 1 X tk X 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: 1 1 X tk X 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 ( ) Xn n + 1 b = 0 k k k=0 MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2 5 for n > 1. Prove that 1 X b t f(t) := n tn = : n! et − 1 n=0 Proof. From the given recurrence, we have − ( ) 1 nX1 n + 1 b = − b n n + 1 k k k=0 for n > 0. Now, we compute the function f(t) as follows: 1 X b tn f(t) = n n! n=0 1 − ( ) X tn nX1 n + 1 = 1 − b (n + 1)! k k n=1 k=0 1 − X nX1 1 = 1 − tn b k!(n + 1 − k)! k n=1 k=0 Put m = n − 1: 1 X Xm 1 = 1 − tm+1 b k!(m + 2 − k)! k m=0 k=0 Interchanging the order of summation, we have: 1 1 X X tm+1 = 1 − b k!(m + 2 − k)! k k=0 m=k Put v = m − k + 2 to get: 1 1 X X tv+k−1 = 1 − b k!v! k k=0 v=2 1 1 X tk−1b X 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).