combinatorial proof of $\sum_{k=1}^{n} k\binom{n}{k}^2 = n\binom{2n-1}{n-1} $

Question

Some one can help me to proof this equation please? I thought about the numbers of ways to choose a group of $n$ from $n$ boys and $n$ girls when the "CEO" must be a boy. But I don't really know how to show it on the left side

Answer 1

That's the way to do it. One the left-hand side, note that $$k\binom{n}{k}^2=k\binom{n}{k}\binom{n}{n-k}$$ so we have the number of ways to choose $k$ boys, times the number of ways to choose $n-k$ girls, times the number of ways to choose which boy is "CEO". Since we must choose at least one boy to be CEO, the sum runs from $1$ to $n$.

Answer 2

$\sum_{k=1}^n k\binom{n}{k}^2 = \sum_{k=1}^n k\binom{n}{k}\binom{n}{n-k}$ Now let $k$ be the number of boys like in your analogy. Now we can choose any number of boys from $1$ to $n$, and then choose $n-k$ girls. Now since a boy must be a "CEO", we have $k$ choices. So the analogies match up perfectly.