Prove that $\sum_{r=1}^n r\binom n r^2 = n\binom{2n-1}{n-1}$ Prove that $\sum_{r=1}^n r\binom n r^2 = n\binom{2n-1}{n-1}$ 
I tried: $\sum_{r=1}^n r\binom n r^2 = n\sum_{r=1}^n \binom {n-1}{r-1}\binom n r$ using the identity $\binom n r = \frac n r \binom {n-1}{r-1}$ for $r \ge 1$.  
Not sure how to proceed from there. I tried expanding the RHS but can't seem to find a way to cancel out the factorials.
 A: So we want to show that
$$\sum^n_{r=1}\binom{n-1}{r-1}\binom{n}{r}=\binom{2n-1}{n-1},$$
or
$$\sum^n_{r=1}\binom{n-1}{n-r}\binom{n}{r}=\binom{2n-1}{n}.$$
Now, how many ways to choose a committee of $n$ people out of $n-1$ males and $n$ females? Consider the cases where there are $r$ females in the committee.
A: A shop has $n$ different doughnuts and $n$ different muffins. I want to choose a total of $n$ items, including at least one doughnut, and I would like to have one doughnut to eat right away, and the remaining $n-1$ put in a bag. How many ways can this be done?
The doughnut I eat right away can be chosen in $n$ ways, and for each of these ways there are $\binom{2n-1}{n-1}$ ways to choose the remaining items.
Let us count another way. If we decide to get $r$ doughnuts, that can be done in $\binom{n}{r}$ ways, and then the doughnut we eat right away can be chosen in $r$ ways. The $n-r$ muffins can be chosen in $\binom{n}{n-r}$ ways, that is, in $\binom{n}{r}$ ways. So there are $r\binom{n}{r}\binom{n}{r}$ to carry out the task while choosing $r$ doughnuts. Add up, $r=1$ to $n$.
A: Using Vandermonde's Identity, we get
$$
\begin{align}
\sum_{r=0}^nr\binom{n}{r}^2
&=\sum_{r=0}^nr\binom{n}{r}\binom{n}{n-r}\\
&=n\sum_{r=0}^n\binom{n-1}{r-1}\binom{n}{n-r}\\
&=n\binom{2n-1}{n-1}
\end{align}
$$
