discrete mathematics - Proof using double counting

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.