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partitions of n into a sum of positive integers

Source: 9th-th Hungary-Israel Binational Mathematical Competition 1998

July 13, 2007
combinatorics proposedcombinatorics

Problem Statement

Let n n be a positive integer. We consider the set P P of all partitions of n n into a sum of positive integers (the order is irrelevant). For every partition α \alpha, let ak(α) a_{k}(\alpha) be the number of summands in α \alpha that are equal to k,k=1,2,...,n. k, k = 1,2,...,n. Prove that αP11a1(α)a1(α)!2a2(α)a2(α)!...nan(α)an(α)!=1. \sum_{\alpha\in P}\frac{1}{1^{a_{1}(\alpha)}a_{1}(\alpha)!\cdot 2^{a_{2}(\alpha)}a_{2}(\alpha)!...n^{a_{n}(\alpha)}a_{n}(\alpha)!}=1.