A combinatorial proof that Σ C(n,k)² = C(2n,n)

Idea. Interpret both sides as the number of ways of choosing $n$ people among $2n$.

Right side. Among $n$ men and $n$ women, the number of committees of $n$ people is $\binom{2n}n$.

Left side. Classify by the number $k$ of men, with $0\le k\le n$. Then there are $n-k$ women. The number of such committees is
$$\binom nk\binom n{n-k}=\binom nk^2,$$
using $\binom n{n-k}=\binom nk$. Summing over $k$ gives the left side.

Check. $n=3$: $1+9+9+1=20=\binom63$.

Generalisation. Vandermonde's identity $\sum_k\binom mk\binom n{r-k}=\binom{m+n}r$.

Lesson. Double counting: to prove an identity, find a set whose size can be computed in two different ways.