Give two proofs that $$ (2n)! = \binom{2n}{n} \cdot (n!)^2 $$
I've already determined how to prove it algebraically (I think):
$\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$
$(2n)! = \frac{(2n)!}{(n!)^2} *(n!)^2$
$(2n)! = (2n)!$
But how would you go about proving it through double counting? Any pointers on how to formulate the proof would be appreciated, thanks.
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