Discrete math: forming a comittee from n men, n women, using 2 different approaches. Prove that,
$$\sum_{k=1}^n k\binom{n}{k}^2 = n\binom{2n-1}{n-1}$$
by determining, in two different ways, the number of ways a committee can
be chosen from a group of $n$ men and $n$ women. Such a committee has a
woman as the chair and has $n − 1$ other members.
Why is $k$ in $[1,n]$? Why is the binomial coefficient to the power $2$? 

We want to select a committee of $n$ people with a chairwoman.
First approach: we choose how many women are there in the committee, say $k$. We choose them among the $n$ women we have in $\binom{n}{k}$ ways. We choose $n-k$ men to complete the committee, among the $n$ we have, in $\binom{n}{n-k}=\binom{n}{k}$ ways. At least, we choose the chairwoman among the selected women. The first approach leads to $\sum_{k=1}^{n}k\binom{n}{k}^2$.
Second approach: we choose the chairwoman ($n$ ways for doing this), then $n-1$ people to complete the committee, from the remaining $2n-1$. This leads to $n\binom{2n}{n-1}$.
Conclusion:
$$\sum_{k=1}^{n}k\binom{n}{k}^2 = n\binom{2n-1}{n-1}.$$
