TITLE: A combinatorial proof of the identity $\sum\limits_{k=0}^nk^2 {n \choose k}^2 = n^2 {2n - 2 \choose n- 1}$
QUESTION [6 upvotes]: I have seen the proof of $\sum_{k=1}^{n} {{k} {n \choose k}^2 ={ n} {{2n-1} \choose {n-1}}}$ done with boys and girls and I somehow understand it, but 
$$\sum_{k=0}^{n}k^2 {n \choose k}^2 = n^2 {2n - 2 \choose n- 1}$$
is a little different.
I tried it like this: 
What is the number of ways a group of $n$ boys and $n$ girls can be divided into a team of $n$ people with a boy leader and a girl supervisor who isn't a part of the team?
On the RHS:
There are two groups of n boys and n girls and we choose the leader from the $n$ boys and then we choose the supervisor from the $n$ girls and then we choose a team of $n - 1$ people for the boy from $2n - 2$ children. 
On the LHS:
We can rewrite it into
$$\sum\limits_{k=0}^nk{n \choose k}k{n \choose n - k}$$
We first choose $k$ boys for a team and then a leader from them and then we choose girls who aren't part of the team. And I don't know what's next.

REPLY [2 votes]: Vandermonde and some basic identities also work:
$$
\begin{align}
\sum_{k=0}^nk^2\binom{n}{k}^2
&=\sum_{k=0}^nk(k-1)\binom{n}{k}^2+\sum_{k=0}^nk\binom{n}{k}^2\tag1\\
&=n(n-1)\sum_{k=0}^n\binom{n-2}{k-2}\binom{n}{n-k}+n\sum_{k=0}^n\binom{n-1}{k-1}\binom{n}{n-k}\tag2\\
&=n(n-1)\binom{2n-2}{n-2}+n\binom{2n-1}{n-1}\tag3\\[3pt]
&=(n-1)^2\binom{2n-2}{n-1}+(2n-1)\binom{2n-2}{n-1}\tag4\\[3pt]
&=n^2\binom{2n-2}{n-1}\tag5
\end{align}
$$
Explanation:
$(1)$: $k^2=k(k-1)+k$
$(2)$: $k(k-1)\binom{n}{k}=n(n-1)\binom{n-2}{k-2}$ and $k\binom{n}{k}=n\binom{n-1}{k-1}$
$(3)$: Vandermonde Identity
$(4)$: $n\binom{2n-2}{n-2}=(n-1)\binom{2n-2}{n-1}$ and $n\binom{2n-1}{n-1}=(2n-1)\binom{2n-2}{n-1}$
$(5)$: $(n-1)^2+(2n-1)=n^2$