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Over at this link, there is a claim that $(2n)! = n!n! {{2n} \choose {n}}$ - see Tom Boardman's answer, the second one down.

I'm wondering why this is the case and if anyone can provide a proof. Is this a special case of a more general relation, and does anyone have any good references for manipulating binomial coefficients and factorials?

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analystic
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1 Answers1

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For integers $a\ge b\ge 0$, we have ${a\choose b}=\frac{a!}{b!(a-b)!}$ (see wiki). Take the special case $a=2b$ and you get ${2b\choose b}=\frac{(2b)!}{b!b!}$, then cross-multiply.

vadim123
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