Competition problem — Combinatorics

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Combinatorics Difficulty 5.9 AIME, harder Prove it

[14.3] Let $m, n$ be non-negative integers. Prove: $\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$ is an integer, with the convention that $0!=1$ .

Solution

None

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.