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 $2$? 
 A: 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 $$\sum_{k=1}^nk\binom nk\binom n{n-k}.$$
A: 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:
$$\sum_{k=1}^{n}k\binom{n}{k}^2 = n\binom{2n-1}{n-1}.$$
A: Another approach not involving combinatorics would be the following. Note that $$\dfrac{k}{n}\binom{n}{k} = \binom{n-1}{k-1}$$
Thus, it's enough to prove that
$$\sum\limits_{k=1}^n  \binom{n-1}{k-1}\binom{n}{k} = \binom{2n-1}{n-1}$$
or, equivalently,
$$\sum\limits_{k=1}^n  \binom{n-1}{k-1}\binom{n}{n-k} = \binom{2n-1}{n-1}$$
But it is true indeed because RHS is the coefficient before $x^{n-1}$ in $(1+x)^{2n-1}$ and LHS is exactly the same, but computed in a different way:
$$(1+x)^{2n-1}=(1+x)^{n-1}(1+x)^n = \left(\sum\limits_{m=0}^{n-1}\binom{n-1}{m}x^{m}\right) \cdot \left(\sum\limits_{m=0}^{n}\binom{n}{m}x^{m}\right)$$
In the expression above it can be seen clearly that the coefficient before $x^{n-1}$ is just a sum of products of coefficients before $x^{k-1}$ and $x^{n-k}$ ($k = 1..n$). That finishes the proof.
