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MATH 630–600. Enumerative Combinatorics
Assignment 4. Due on Wednesday, October 22, 2008
- Let D(n) be the number of derangement of length n. Prove that the generating function of D(n) satisfies ∑
n≥ 0
D(n)xn n!
e−x (1 − x)
- Let m > n be two positive integers. How many ballot sequences (a 1 , a 2 ,... , am+n) such that
(a) ai ∈ {+1, − 1 }, (b) a 1 + a 2 + · · · + ai > 0 for all i = 1, 2 ,... , m + n, (c) a 1 +a 2 +· · ·+am+n = m−n. That is, the multiset {a 1 ,... , am+n} = {(+1)m, (−1)n}.
- Given a sequence (an)n≥ 0 , let A(x) =
n≥ 0 anx n (^) be its generating function. Suppose that the sequence (an) is the binomial transform of a sequence (bn), i.e.,
an =
∑^ n
i=
n i
bi ∀n ≥ 0.
(a) Show that A(x) = (^1) −^1 x B( (^1) −xx ). (b) Deduce that B(x) = (^) 1+^1 x A( (^) 1+xx ). (c) Deduce from (b) that
bn =
∑^ n
i=
(−1)n−i
n i
ai, ∀n ≥ 0.
This is another proof of the binomial inversion formula.
- A sequence of positive integers (a 1 , a 2 ,... , an) is said to be of restricted growth if a 1 = 1, ai+1 ≤ 1 + max{a 1 ,... , ai}, for all i = 1, 2 ,... , n. (a) Write all restricted growth sequences for n = 4. (b) Show that the number of restricted growth sequences is the Bell number Bn. (c) Describe the Stirling numbers S(n, k) in terms of restricted growth sequences.
- Show by a combinatorial argument that
(a) (^) ( n n − k
q
= qk(n−k)
n k
q−^1
where
(n k
q−^1 is obtained from^
(n k
q by replacing^ q^ with^ q
(Hint: use lattice paths from (0.0) to (n − k, k)). (b) Deduce that
(n k
q is a symmetric polynomial of^ q, that is, if ( n k
q
= a 0 + a 1 q + a 2 q^2 + · · · + aN qN
with aN 6 = 0, then ai = aN −i for all i.
- Vandermonde’s formula for the q-binomial coefficients is ( n + m p
q
∑^ p
k=
q(m−k)(p−k)
m k
q
n p − k
q
Prove this formula by a lattice path counting argument. (Hint: Count lattice paths from (0, 0) to (p, m + n − p), and consider the intersection between such paths with the line x + y = n. )
- Exercise 1 on textbook, page 86.
- Exercise 2a on textbook, page 87.
- Exercise 7 on textbook, page 88.
