Download e-book for iPad: An Introduction to Combinatorics and Graph Theory by David Guichard

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Show that x = n−1 k=0 (x − k) = i=0 s(n, i)x , n ≥ 1; x is called a falling factorial. Find n a similar identity for xn = n−1 k=0 (x + k); x is a rising factorial. n n xk = xn , n ≥ 1; xk is defined in the previous exercise. The previous k k=0 exercise shows how to express the falling factorial in terms of powers of x; this exercise shows how to express the powers of x in terms of falling factorials. 6. Show that n−1 7. Prove: S(n, k) = i=k−1 8. Prove: n = k n−1 S(i, k − 1). i n−1 (n − i − 1)! i=k−1 n−1 i i .

N−1 n−1 i=0 i=0 Now replacing x by −x gives −n (1 − x) ∞ = i=0 −n So (1 − x) is the generating function for 1, ∞ · 2, . . , ∞ · n} of size i. n+i−1 i x. 1 Newton’s Binomial Theorem 53 In many cases it is possible to directly construct the generating function whose coefficients solve a counting problem. 3 Find the number of solutions to x1 + x2 + x3 + x4 = 17, where 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 5, 0 ≤ x3 ≤ 5, 2 ≤ x4 ≤ 6. We can of course solve this problem using the inclusion-exclusion formula, but we use generating functions.

3 Find the number of solutions to x1 + x2 + x3 + x4 = 17, where 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 5, 0 ≤ x3 ≤ 5, 2 ≤ x4 ≤ 6. We can of course solve this problem using the inclusion-exclusion formula, but we use generating functions. Consider the function (1 + x + x2 )(1 + x + x2 + x3 + x4 + x5 )(1 + x + x2 + x3 + x4 + x5 )(x2 + x3 + x4 + x5 + x6 ). We can multiply this out by choosing one term from each factor in all possible ways. If we then collect like terms, the coefficient of xk will be the number of ways to choose one term from each factor so that the exponents of the terms add up to k.

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An Introduction to Combinatorics and Graph Theory by David Guichard


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