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.
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 ${2n \choose n-1}$.