Prove the identity $\frac{(2n)!}{2^n \cdot n!} = (2n-1)(2n-3)\cdots 3\cdot 1$ using combinatorial proof.

Question

Prove the identity $$\frac{(2n)!}{2^n \cdot n!} = (2n-1)(2n-3)\cdots 3\cdot 1$$ using combinatorial proof.

Answer 1

Hint. It counts the number of ways to choose $n$ disjoint pairs of objects from $2n$ objects.

Answer 2

Ok, I think, I get the LHS: we order $2n$ objects, thus $(2n)!$, we don't care about the order of objects in each pair, thus $2^n$, and we don't about the order of pairs, thus $n!$ I am struggling with the RHS though. Looks like proof by induction, but I don't see it...

EDIT: I think, you could apply recursive induction for RHS. There $2n-1$ possibilities to pick the first pair from $2n$ objects times number of ways to pick $n-1$ distinct pairs out of $2n-2$ objects.