Combinatorial identity using combinatorial argument: $\left[\sum\limits_{k=0}^{n} \binom nk\right] ^ 2 = \sum\limits_{k=0}^{2n}\binom{2n}k$

Suppose that you have a group of $n$ mixed couples. The lefthand side is the number of ways to choose a set of $k$ men and a set of $\ell$ women for some $k,\ell\in\{0,\dots,n\}$; the righthand side is the number of ways to choose an arbitrary subset of the group. Clearly these are the same.

One can easily give a combinatorial argument for that identity, but the understanding of the situation is not complete without observing that the identity is immediate from the auxiliary formulas $$ \sum_{k=0}^m \binom mk = 2^m \quad\text{for all $m\in\mathbf N$, and}\qquad (2^n)^2 = 2^{2n}, $$ both of which are obvious and have obvious combinatorial proofs, via $$ \left[\sum_{k=0}^n \binom nk\right]^2=(2^n)^2 = 2^{2n} =\sum_{k=0}^{2n} \binom {2n}k. $$ If you compose the bijection corresponding to the first identity above for $m=n$ (which involves double counting of the subsets of an $n$-set), with the one for $(2^n)^2 = 2^{2n}$ (which involves double counting of the subsets of the disjoint union of two $n$-sets), and then again with the bijection for the first identity but for $m=2n$ and in the opposite direction, then you get a bijection for the outer identity given in the question. Indeed this is exactly the bijection described in the answers by Brian Scott and anon.