Vandermonde's identity

Statement

Let $m$, $n$ and $r$ be natural numbers. Then $$\binom{m + n}{r} = \sum_{k=0}^{r} \binom{m}{k} \binom{n}{r - k}.$$

Remarks

Combinatorial proof: take a set of $m$ men and $n$ women. An $r$-element committee has some number $k$ of men and $r - k$ women. With $m = n = r$, and the symmetry of binomial coefficients, it gives $\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}^2$.

Explanations

L1 human

To choose $r$ people from two groups together, choose $k$ from the first group and $r - k$ from the second, for each possible $k$. Then add the counts for all values of $k$.