Help with combinatorial proof of binomial identity: $\sum\limits_{k=1}^nk^2{n\choose k}^2 = n^2{2n-2\choose n-1}$

$$\sum_{k=1}^nk^2\binom{n}k^2=n^2\binom{2n-2}{n-1}\;.\tag{1}$$

You have $n$ men and $n$ women. You want to choose a team of $n+1$ people, and on this team you will designate a man and a woman as co-captains. You can choose the co-captains in $n^2$ ways, and you can then choose the remainder of the team in $\binom{2n-2}{n-1}$ ways, so the righthand side of $(1)$ counts the possible teams.

Now let’s count the teams with exactly $k$ women, for $k=1,\dots,n$. There are $\binom{n}k$ ways to choose the women, and $k$ ways to designate one the female co-captain. There are then $n$ ways to pick the male co-captain and $\binom{n-1}{n-k}$ ways to pick the remaining men. And

$$kn\binom{n}k\binom{n-1}{n-k}=kn\binom{n}k\binom{n-1}{k-1}=k^2\binom{n}k^2\;,$$

so the lefthand side counts the same thing according to the number of women on the team.

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We'll use ( several times ! ) the identity $\ds{% {m \choose \ell} = \oint_{\verts{z} = 1}{\pars{1 + z}^{m} \over z^{\ell + 1}} \,{\dd z \over 2\pi\ic}}$ which provides a systematic evaluation in several cases.

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\begin{align} \color{#f00}{\sum_{k = 1}^{n}k^{2}{n \choose k}^{2}} & = \sum_{k = 0}^{n}k^{2}{n \choose k}{n \choose n - k} = \sum_{k = 0}^{n}k^{2}{n \choose k} \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n - k + 1}} \,{\dd z \over 2\pi\ic} \\[3mm] &= \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}} \sum_{k = 0}^{n}{n \choose k}k^{2}z^{k}\,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}}\bracks{% \pars{z\,\partiald{}{z}}^{2} \sum_{k = 0}^{n}{n \choose k}z^{k}}\,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}}\bracks{% \pars{z\,\partiald{}{z}}^{2}\pars{1 + z}^{n}}\,{\dd z \over 2\pi\ic} \\[3mm] & = \oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n + 1}}\bracks{% n\pars{1 + z}^{n - 2}\pars{nz^{2} + z}}\,{\dd z \over 2\pi\ic} \\[3mm] & = n^{2}\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{\pars{n - 2} + 1}} \,{\dd z \over 2\pi\ic} + n\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 2} \over z^{\pars{n - 1} + 1}} \,{\dd z \over 2\pi\ic} \\[3mm] & = n^{2}{2n - 2 \choose n - 2} + n{2n - 2 \choose n - 1} \\[3mm] & = n^{2}\,{\pars{2n - 2}! \over \pars{n - 2}!\,n!} + n\,{\pars{2n - 2}! \over \pars{n - 1}!\pars{n - 1}!} = n\,{\pars{2n - 2}! \over \pars{n - 1}!} \bracks{{n - 1 \over \pars{n - 1}!} + {1 \over \pars{n - 1}!}} \\[3mm] & = \color{#f00}{n^{2}{2n - 2 \choose n - 1}} \end{align}

This is just a way to reformulate the answer by Brian Scott so that it avoids any algebraic manipulation of binomial coefficients.

Consider colouring the squares of a $2\times n$ rectangle with colours red, white, blue, subject to the following restrictions

• Exactly one square of each of the two rows must be white,
• There must be $n-1$ each of red and blue squares.

Every solution has equally many blue squares in the first row as red squares in the second row; let this number be $n-k$ (with $1\leq k\leq n$). Now for any solution with this value of $k$ we must choose in the first row $k-1$ red squares, $1$ white square and $n-k$ blue squares, and in the second row similarly with red and blue interchanged. For each row this number is given by the trinomial coefficient $\binom n{k-1,1,n-k}$, giving $$ \sum_{k=1}^n\binom n{k-1,1,n-k}^2 $$ solutions in all. On the other hand, one may just choose the white squares in both rows (giving the factor $n^2$), and then the $n-1$ red squares among the remaining $2(n-1)$, for $$ n^2\binom{2(n-1)}{n-1} $$ possibilities. Finally the trinomial coefficient above can be seen to be $k\binom nk$: for $\binom nk$ ways to choose the subset for the first two colours together, and $k$ choices for the single white square among them.