Give a combinatorial proof that $\sum_{k=1}^{n} {{k} {n \choose k}^2 ={ n} {{2n-1} \choose {n-1}}}$

Question

$$\sum_{k=1}^{n} {{k} {n \choose k}^2 ={ n} {{2n-1} \choose {n-1}}}$$

How would I approach this problem to make a combinatorial proof?

Answer 1

We have $2n$ politicians, $n$ of them belonging to the right wing, the other $n$ belonging to the left wing. We have to select a committee made by $n$ politicians and nominate one of its member in the left wing as chief of the committee. In how many ways can we choose such a committee? There are two ways of counting that. We may select the chief first ($n$ possible choices out of the $n$ people in the left wing), then the other members (we have to select $n-1$ people among $2n-1$). Or, for any $k$ ranging from $1$ to $n$, we may choose $k$ people in the left wing to be part of the committee, $k$ in the right wing not belonging to the committee, then nominate the chief. That argument proves: $$ \sum_{k=0}^{n}\binom{n}{k}\binom{n}{k}k = n\binom{2n-1}{n-1}.$$

A very small variation about $\sum_{k=0}^{n}\binom{n}{k}^2=\binom{2n}{n}$ (see here), as pointed in the comments.