# Binomial sum which adds to $2^n n!$

I'm looking for a combinatorial interpretation for the identity $$\sum_{k=0}^n\binom nk (2k-1)!!\,(2n - 2k - 1)!! = 2^n n!$$ where $$(2n - 1)!! = (2n - 1)(2n - 3) \cdots 5 \cdot 3 \cdot 1$$.

Perhaps the most natural interpretation of the right-hand side is the number of $$2$$-colorings of the letters of the permutations on $$[n] = \{1,2,...,n\}$$. However, I can't find a way to make the sum fit this interpretation.

Perhaps there is a way to use the fact that $$(2n-1)!!$$ is the number of ways to choose $$n$$ disjoint pairs of items from $$2n$$ items?

We can do some trickery to show that this is equivalent to showing that $$\sum_{k=0}^n\binom{2n}{n}\binom{2n-2k}{n-k} = 4^n,$$ which is addressed by this question, but I'm interested in the earlier interpretation.