Using generating functions show that $\sum^n _{r=1}r\binom{n}{r}\binom{m}{r} =n\binom{n+m-1}{n}$ 
Using generating functions show that $$\sum^n _{r=1}r\binom{n}{r}\binom{m}{r} =n\binom{n+m-1}{n}$$

I have been thinking about it as follows. The RHS is a coefficient of $x^n$ in the expansion of $$\frac{mx}{(1-x)^{m+1}}$$ The LHS is problematic though. It must be a coefficient of a product of some generating functions. The fact that we have binomial coefficients suggests applying the binomial theorem maybe, but I have not figured out how to do that in this case.
 A: I would prefer complex variables for this one but it can be done using generating functions only. We have for the LHS that it is
$$\sum_{r=1}^n n {n-1\choose r-1} {m\choose r}
= n \sum_{r=0}^{n-1} {n-1\choose r} {m\choose r+1}.$$
Note that when $m-1\gt n-1$ we can extend $r$ to $m-1$ because the first binomial coefficient is zero then. And when $m-1\lt n-1$ or $m\lt n$ we can reduce the sum to $m-1$ because the second binomial coefficient is zero then.
This yields
$$n \sum_{r=0}^{m-1} {n-1\choose r} {m\choose r+1}.$$
Now this is
$$n\sum_{r=0}^{m-1} [z^r] (1+z)^{n-1} [z^{m-r-1}] (1+z)^m.$$
which is precisely
$$n [z^{m-1}] (1+z)^{n-1} (1+z)^m = n [z^{m-1}] (1+z)^{n-1+m}
= n {n-1+m\choose m-1} = n {n+m-1\choose  n}.$$
So the two generating functions we were looking for are $(1+z)^{n-1}$ and $(1+z)^m.$
Addendum. In terms of complex variables we put
$${m\choose r+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m-r}} (1+z)^m \; dz$$
which yields for the sum
$$\frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m}} (1+z)^m 
\sum_{r=0}^{n-1} {n-1\choose r} z^r\; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m}} (1+z)^m 
(1+z)^{n-1}\; dz
\\ = \frac{n}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m}} (1+z)^{m+n-1}\; dz.$$
This evaluates to $$n{m+n-1\choose m-1}$$ by inspection.
