By David Guichard

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**Extra resources for An Introduction to Combinatorics and Graph Theory [Lecture notes]**

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Rn−1 = a + (n − 1)m mod n. Label n − 1 boxes with the numbers 0, 1, 2, 3, . . , b − 1, b + 1, . . n − 1. Put each ri into the box labeled with its value. Two remainders end up in the same box, say ri and rj , with j > i, so ri = rj = r. This means that a + im = q1 n + r Hence and a + jm = q2 n + r. a + jm − (a + im) = q2 n + r − (q1 n + r) (j − i)m = (q2 − q1 )n. Since n is relatively prime to m, this means that n | (j − i). But since i and j are in {0, 1, 2, . . , n − 1}, 0 < j − i < n, so n | (j − i).

Let pk (n) be the number of partitions of n into exactly k parts. We will find a recurrence relation to compute the pk (n), and then n pn = pk (n). k=1 Now consider the partitions of n into k parts. Some of these partitions contain no 1s, like 3 + 3 + 4 + 6, a partition of 16 into 4 parts. Subtracting 1 from each part, we get a partition of n − k into k parts; for the example, this is 2 + 2 + 3 + 5. The remaining partitions of n into k parts contain a 1. If we remove the 1, we are left with a partition of n − 1 into k − 1 parts.

I! (n + i − 1)! = (−1)i i! (n − 1)! n+i−1 n+i−1 = (−1)i = (−1)i . i n−1 Thus ∞ ∞ n+i−1 −n i n+i−1 i (x + 1) = (−1) x = (−x)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.