Vandermonde's identity and a weighted sum of squares

1. Count the committees of $r$ people chosen among $m$ men and $n$ women in two ways. Directly: $\binom{m+n}r$. By the number $k$ of men: choose $k$ men in $\binom mk$ ways and $r-k$ women in $\binom n{r-k}$ ways. Summing over $k$ gives the identity.

2. Use $k\binom nk=n\binom{n-1}{k-1}$. Then
$$\sum_{k=0}^nk\binom nk^2=n\sum_{k=1}^n\binom{n-1}{k-1}\binom nk=n\sum_{k=1}^n\binom{n-1}{k-1}\binom n{n-k}.$$
In the last sum the lower indices add up to $(k-1)+(n-k)=n-1$, so with $j=k-1$ and Vandermonde (with $m=n-1$, $n$, $r=n-1$) we get $\sum_{j}\binom{n-1}{j}\binom{n}{n-1-j}=\binom{2n-1}{n-1}$. Hence the sum equals $n\binom{2n-1}{n-1}$.

Check. $n=3$: $1\cdot3^2+2\cdot3^2+3\cdot1^2=30=3\binom52$ ✓.

Lesson. To prove a binomial identity, find a set counted in two ways; extract factors with $k\binom nk=n\binom{n-1}{k-1}$ to reduce to Vandermonde.