MATH 689. Enumerative Combinatorics
Solution of Assignment 2.
1. Let A(n, k) be the Eularian number, that is, A(n, k) is the number of permutations of length
nwith k−1 descents. Prove that A(n, k) satisfies the recurrence
A(n, k) = kA(n−1, k )+(n−k+ 1)A(n−1, k −1),for 2 ≤k≤n,
with boundary conditions A(n, 0) = 0, A(n, 1) = A(n, n) = 1 and A(n, k) = 0 for k > n.
Solution. For a permutation π=a1a2. . . an∈ Sn, assume n=ai.
Case 1. If i=n, or i6=nand ai−1> ai+1, then removing nwill yield a permutation of length n−1
with k−1 descents. There are A(n−1, k ) such permutations in Sn−1. Conversely, for each such a
permutation, one can insert nback in any of the descent position, or at the position n. There are
ksuch choices. Therefore, kA(n−1, k ) permutations in Snfall into this case.
Case 2. If i6=nbut ai−1< ai+1, then removing nwill yield a permutation of length n−1 with
k−2 descents. Conversely, for each such a permutation, one can insert ninto the first position, or
any of the non-descent positions. There are (n−1) −(k−2) = n−k+ 1 choices. Hence there are
(n−k+ 1)A(n−1, k −1) permutations in Snfall into this case.
2. Again A(n, k) is the Eularian number. Prove that
A(n, k) = A(n, n + 1 −k).
Solution. If a1a2· · · anhas k−1 descents, then anan−1· · · a2a1has n−1−(k−1) = n−kdescents.
3. Let S⊆[n−1], and let α(S) denote the number of n-permutations whose descent set is
contained in S.
Find the one-element set {i} ⊆ [n−1] for which α({i}) is maximal.
Solution. α({i}) = n
i. Hence the maximal is reached when i=bn/2cand dn/2e.
4. Prove that for any permutation π,i(π) = i(π−1), where i(π) is the number of inversions of
π.
Solution. If (π(i), π(j)) is an inversion of πiff i < j and π(i)> π (j). But this is exactly same as
positions π(j)< π(i), and π−1(π(j)) > π−1(π(i)).
5. A permutation πis called even if i(π) is even. Similarly, πis called odd if i(π) is odd. Let
n≥2. Prove that the number of even (odd) permutations of length nis n!/2.
Solution. For any π∈ Snwith π=a1a2a3· · · an, let f(π) be the permutation obtained from π
by exchanging the first two entries, i.e., f(π) = a2a1a3· · · an. Then fis an involution of Snwhich
maps even permutations to odd permutations. Hence the number of even permutations is the same
as the number of odd permutations, and is equal to n!/2.
6.Exercises on page 49, Problem 31. For a permutation π, let m(π) denote the number of
left-to-right maxima of πand i(π) the number of inversions of π. Compute the generating function
F(x, q) = X
π∈Sn
xm(π)qi(π).
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