MATH 580, midterm exam SOLUTIONS

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1 MATH 580, midterm exam SOLUTIONS Solve the problems in the spaces provided on the question sheets. If you run out of room for an answer, continue on the back of the page or use the Extra space empty page at the end. Answers without proofs or incomplete proofs will not receive full credit. NO notes, books, or any electronic devices are allowed. Name: Penn ID number: Run L A TEX again to produce the table

2 1. (10 points) For any positive integer n, let f(n) be the number of integer partitions λ of n, such that each part size appears exactly 0, 2, 3 or 5 times. E.g. λ (9, 9, 6, 6, 6, 6, 6, 2, 2, 2, 1, 1) is such a partition. Set f(0) 1. Let F (q) f(n)q n be the generating function for these numbers. (a) Find a product type formula for F (q). n0 Solution: Let λ (1 m1 2 m2...), i.e. the number of parts of λ equal to i is m i. In this case we have that m i can only take the values 0, 2, 3, 5. Then i1 F (q) λ m i {0,2,3,5} q λ q mii (m 1,m 2,...) m i {0,2,3,5} q i imi ( 1 + q 2i + q 3i + q 5i) i1 (1 + q 2i )(1 + q 3i ) (b) Prove that f(n) is equal to the number of partitions µ (µ 1, µ 2,...) of n, such that µ i 2 (mod 4) or µ i 3 (mod 6) for each i. Solution: Using the formula from part (a), we have f(n)q n + q n i1(1 2i )(1 + q 3i ) i i i1 (1 + q 2i ) (1 q2i ) (1 q 2i ) (1 + q3i ) (1 q3i ) (1 q 3i ) (1 q 4i )(1 q 6i ) 1 q 2i )(1 q 3i ) i(1 q4i )(1 q 6i ) j (1 q2j )(1 q 3j ) 1 j:odd (1 q2j )(1 q 3j ), where the last equality comes from the cancelation of the numerator with all terms (1 q 2j )(1 q 3j ) with j even. The last line is the generating function for partitions whose parts are of the form 2j or 3j for j 2r +1 odd, i.e. 2(2r +1) 4r +2 2 (mod 4), or 3(2r +1) 6r +3 3 (mod 6). Page 2

3 2. (10 points) Let A n (x) be the Eulerian polynomial, i.e. A n (x) w S n x des(w)+1. Prove that A n (2)/2 is equal to the number of ordered set partitions of [n] (i.e. sequences (S 1, S 2,...), s.t. S i [n], S i, i S i [n] and S i S j for i j). Solution: This is HW problem 1.133a. We have that A n (2)/2 w S n 2 des(w). Let T {(w, D) w S n, D D(w)}, i.e. pairs of permutations w and a subset S of its descents. Then #T A n (2)/2. Construct the following bijection φ between T and the set O of ordered set partitions: Let A [n 1] \ D(w), i.e. the set of ascents of w and let A D {i 1,..., i r } ordered in increasing order (i.e. i 1 < i 2 <... are the indices where w has an ascent or a descent from D). Let i 0 1. Then let φ(w, D) : (S 1, S 2,...), where S p+1 {w ip+1, w ip+2,..., w ip+1 } for p 0, 1,.... In other words, S p+1 is the set of elements of w between positions i p and i p+1. Note that since none of i p + j that appear between i p and i p+1 is an ascent, so we necessarily have w ip+1 > w ip+2 > > w ip+1. We show that φ is a bijection by constructing its inverse: Let (S 1,..., S k ) be a set partition of [n] and let S i {a i 1,..., a i r i }, writing its elements in a decreasing order, i.e. a i 1 > a i 2 >. Let w a 1 1,..., a 1 r 1, a 2 1,..., a 2 r 2, a 3 1,... be the permutation obtained by concatenating the sequences a i in this order. Finally, having constructed w, let D be the set of indices from R {r 1, r 1 + r 2, r 1 + r 2 + r 3,...}, where w has a descent, i.e. w r1+r 2+ +r i a i r i > a i+1 1 w r1+r 2+ +r i+1. To verify that φ(w, D) S, we note that clearly D D(w) and A R \ D, since there are no ascents between elements a i j, ai j+1, as each sequence ai is decreasing. Page 3

4 3. (10 points) Let k 1 be fixed. Let f k (n) be the number of permutations of [n], which have NO cycle of length k. (E.g. if k 3 and n 6, then (1, 5)(2, 4, 3, 6) has no cycle of length 3, whereas (1, 5, 3)(4, 2, 6) has two such cycles and should not be considered.) Prove that f k (n) n/k i0 ( 1) i n! i!k i. Solution: This is HW problem For every c (c 1,..., c k ) cycle of length k in S n, let A c be the set of permutations whose cycle decomposition contains c. Then f k (n) #S n \ ( c A c ) n! # ( c A c ). By the inclusion-exclusion formula we have # ( c A c ) A c A c 1 A c 2 + c {c 1,c 2 } If c 1 c 2, then A c 1 A c 2 since any element of a permutation is part of a unique cycle (in the unique cycle factorization), there can be no element part of two different cycles. Thus for any i cycles c 1,..., c i, we have that A c 1 A c i if c r c p for some r p i. Else, when they are all pairwise disjoint, we have that A c 1 A c i {w c 1 c 2 c i u}, where u is any permutation of the elements in [n] \ (c 1 c 2...) (i.e. the remaining n ki elements). Thus A c 1 A c i (n ki)! The number of i-tuples of disjoint cycles (c 1, c 2,..., c i ) is determined by choosing k disjoint elements for each cycle in a total of ( n k,k,...,k,n ki) ways and then arranging those elements in each cycle in k!/k (k 1)! ways. This gives a total number of ( ) n ((k 1)!) i n! k, k,..., k, n ki k i (n ki)! ways. However, we need the unordered i-tuples, so we divide by i!, as each tuple can be arranged in i! ways. Finally, putting all this together in the inclusion-exclusion formula, we get # ( c A c ) A c A c 1 A c 2 + c {c 1,c 2 } n/k ( 1) i n! 1 k i (n ki)! i! A c 1 A c i ( 1) i n! 1 k i (n ki)! (n ki)! i! i i i0 ( 1) i n! i!k i Page 4

5 4. (15 points) (Bonus problem**) Let T (n) be the set of permutations w S n, which, when written in line notation w w 1 w 2... w n, satisfy: ( ) there are NO indices 1 i < j < n, s.t. w i < w j < w j+1. Show that #T (n) B(n), where B(n) are the Bell numbers the number of unordered set partitions of [n]. Solution: Bijective proof: Let the set of such permutations be A n, we construct the following bijection ψ from A into the set B of set partitions of [n]: Let w A and let i be the position of 1 in w, i.e. w i 1. Since w i < w j for all j i + 1,..., n, we must have that w j > w j+1 for all j > i, as otherwise 1 w i < w j < w j+1, contradicting condition ( ) for w. Let S 1 {1, w i+1,..., w n } be the set of the elements of w with indices i, i + 1,..., n. Next, let w 2 w 1... w i 1 be the remaining permutation from w, let i 1 < i be the index of its smallest element, i.e. w i1 w j for all j {1,..., i 1}. Since w 2 also satisfies ( ), we have that w i1+1 > w i1+2 >... > w i 1 > w i1. Let S 2 {w i1, w i1+1,..., w i 1 } be the set of these elements. Then construct w 3 w 1..., w i1 1, and get S 3 similarly as before and so on. Let ψ(w) {S 1, S 2,...}. It is clear by construction that this is a set partition. To show that ψ is a bijection, we construct its inverse: Given a set partition {S 1,...}, arrange the sets in order of their minimal elements to get (S 1, S 2,...), so necessarily we have 1 S 1. Let S 1 \ {1} {a 1 > a 2 >... > a r } (i.e. arrange its elements in decreasing order), then we let the last r + 1 elements of the permutation w be 1a 1... a r. Take then S 2, let a 2 0 be its minimal element, arrange the remaining elements in decreasing order a 2 1 > a 2 2 > and glue a 2 0a at the beginning of w. Continue the same way with S 3 etc. It remains to show that the resulting permutation w indeed satisfies ( ). Suppose not, then there exist i < j, s.t. w i < w j < w j+1. The only places where w has ascents is between a r 0 and a r 1 for some r. (Since a r 0 is minimal in S r and also smaller than any a p 0 with p > r by the order we chose for the sets S, we cannot have a r 0 > a r+1 j for any r and j.) But then w j a r 0, and w i < a r 0 with w i S p for some p > r, which contradicts the minimality of a r 0 among S r, S r+1,.... Recursive proof: We have that the numbers B(n) satisfy the following easy recurrence: B(n + 1) n k0 ( ) n B(k). k The same recursion holds for A(n) #A n by establishing a bijection between A n+1 and k {(u, S) u A n k, S [n], #S k}. Let w A n+1, let w i 1, and S {w i+1,..., w n }. Let u w 1... w i 1, i.e. u is the permutation of [i 1] whose elements are arranged in the same relative order as the ones in w 1... w i 1. Clearly u also satisfies ( ). Conversely, given such u and S, construct w ũ1s >, where S > is the elements of S + 1 arranged in decreasing order and ũ is the permutation of [n + 1] \ {1} S in the same order as u. Then w satisfies ( ): otherwise, if w j < w j+1, we must have either w j 1, in which case there is no i with w i < w j 1, or else j + 1 < i, in which case all these entries are in ũ, so u doesn t satisfy ( ). Putting all this together, we have Page 5

6 A(n + 1) k #{(u, S) u A n k, S [n], #S k} k ( ) n A(n k) k and with B(1) A(1) 1 we get that both satisfy the same recursion and initial conditions, soa(n) B(n). Page 6

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