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 $2$? 

Hint: $k$ is the total number of women on the committee (so must be at least $1$). It may help to use the fact that $\binom nk=\binom n{n-k}$ to rewrite the LHS as $$\sum_{k=1}^nk\binom nk\binom n{n-k}.$$
