The double-counting proof of a binomial identity

Consider a set of $2n$ people consisting of $n$ boys and $n$ girls. We count, in two ways, the number of ways to choose a committee of exactly $n$ people from this set of $2n$.

Directly, by definition, this number is $\binom{2n}{n}$, since we are simply choosing $n$ objects out of $2n$.

Alternatively, split every valid committee according to how many boys it contains. If the committee contains exactly $k$ boys for some $k$ with $0 \le k \le n$, then those $k$ boys can be chosen in $\binom{n}{k}$ ways, and the remaining $n-k$ committee members must be girls, chosen from the $n$ girls in $\binom{n}{n-k}$ ways. By the symmetry identity $\binom{n}{n-k} = \binom{n}{k}$, the number of committees with exactly $k$ boys is $\binom{n}{k} \cdot \binom{n}{k} = \binom{n}{k}^2$.

Every valid committee of $n$ people has some well-defined number of boys $k$ between $0$ and $n$, and no committee is counted twice across different values of $k$, so summing over all $k$ gives the total number of committees as $\sum_{k=0}^{n} \binom{n}{k}^2$.

Since both expressions count exactly the same set of committees, they must be equal: $\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$.