By David Guichard

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N k or n k will contain 11. Under the assumptions that s(n, 0) = s(0, n) = 0 for n = 0, and s(n, k) = s(n − 1, k − 1) − (n − 1)s(n − 1, k), extend the table for s(n, k) to all integers, and find a connection to S(n, k) similar to that in the previous problem. 12. 4. 13. 6. 1 How many submultisets of the multiset {2 · a, 4 · b, 3 · c} have size 7? We recast the problem: this is the number of solutions to x1 + x2 + x3 = 7 with 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 4, 0 ≤ x3 ≤ 3. We know that the number of solutions in nonnegative integers is 7+3−1 = 92 , so this is an overcount, since we count solutions that 3−1 do not meet the upper bound restrictions.

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.

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. This gives us a 1–1 correspondence between the partitions of n into k parts, and the partitions of n − k into k parts together with the partitions of n − 1 into k − 1 parts, so pk (n) = pk (n − k) + pk−1 (n − 1). Using this recurrence we can build a triangle containing the pk (n), and the row sums of this triangle give the partition numbers.

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An Introduction to Combinatorics and Graph Theory [Lecture notes] by David Guichard

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