MathDB
A 26

Source:

May 25, 2007
floor functioninequalitiesinductionratioDivisibility Theory

Problem Statement

Let mm and nn be arbitrary non-negative integers. Prove that (2m)!(2n)!m!n!(m+n)!\frac{(2m)!(2n)!}{m! n!(m+n)!} is an integer.
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