MathLabs

Problem 3

Let mm and nn be arbitrary non-negative integers. Prove that (2m)!(2n)!m!n!(m+n)!\dfrac{(2m)!(2n)!}{m!n!(m+n)!} is an integer. (0!=10!=1).
Step 1 of 4: Define the target ratio
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In plain words

Naming the expression exposes a useful recurrence.

f(m,n)=(2m)!(2n)!m!n!(m+n)!f(m,n)=\dfrac{(2m)!(2n)!}{m!n!(m+n)!}
Detailed analysis

Let f(m,n)f(m,n) denote the displayed factorial ratio. We prove f(m,n)f(m,n) is integral for all non-negative m,nm,n.