Combinatorial identity $\sum\limits_{k=0}^{n}\frac{n-k}{k+1}\binom{n}{k}^2 = \binom{2n}{n-1}$ I have an identity $$\sum_{k=0}^{n}\frac{n-k}{k+1}\binom{n}{k}^2 = \binom{2n}{n-1}$$ for which I'm looking for a combinatorial proof. Any ideas? 
I was thinking about separating $2n$ on boys and girls, but the fraction that appears on the LHS seems problematic. Dividing by $k+1$ suggests some element being chosen on $k+1$ ways, but I don't see what could be a possible story to that.
 A: One way to deal with the left-hand side is to notice that
$$
\frac{n-k}{k+1} {n \choose k} = {n \choose k+1} = {n \choose n-1-k}.
$$
Therefore the left-hand side is 
$$
\sum_{k=0}^{n-1} {n \choose k}{n \choose n-1-k}.
$$
One way to look at this sum is that it counts the number of ways to select a group of size $n-1$ from a population of $n$ girls and $n$ boys. Why? Suppose that we want to count the ways to do this task with the added condition that we choose $k$ girls. So there are ${ n\choose k}$ ways to pick the $k$ (out of $n$) girls and ${n \choose n-1-k}$ ways to pick the remaining $n-1-k$ (out of $n$) boys for the group. Summing over all possible $k$ completes this task. 
Now clearly, the number of ways to select a group of size $n-1$ from a population of $2n$ people is ${2n \choose n-1}$. 
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{k = 0}^{n}{n - k \over k + 1}{n \choose k}^{2}} & =
\sum_{k = 0}^{n}{n \choose k}{\color{#f00}{n} \choose \color{#f00}{n - k - 1}} =
\sum_{k = 0}^{n}{n \choose k}
\oint_{\verts{z} = 1}{\pars{1 + z}^{\color{#f00}{n}} \over z^{\color{#f00}{n - k -1} + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n}}
\sum_{k = 0}^{n}{n \choose k}z^{k}\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n} \over z^{n}}
\pars{1 + z}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\oint_{\verts{z} = 1}{\pars{1 + z}^{\color{#f00}{2n}} \over
z^{\color{#f00}{n - 1} + 1}}\,{\dd z \over 2\pi\ic} =
\color{#f00}{2n \choose n - 1}
\end{align}
