Proof using Generating Functions I need to use generating functions to show that
$$\sum_{r=1}^n r\binom{n}{r}\binom{m}{r} = n\binom{n+m-1}{n}$$
I have the generating functions for the right-hand side, which is
$$\frac{mx}{(1-x)^{m+1}}$$
Then the coefficient of $x^n$ is $n\binom{n+m-1}{n}$, but I do not know how to proceed for the left-hand side.
 A: Here  is  a variation with $$(1+x)^{n+m-1}=\sum_{k=0}^{n+m-1}\binom{n+m-1}{k}x^k$$ as generating function. It is convenient to use the coefficient of  operator  $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write for instance
\begin{align*}
\binom{n+m-1}{n}=[x^n](1+x)^{n+m-1}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{r=1}^nr\binom{n}{r}\binom{m}{r}}
&=m\sum_{r=1}^n\binom{n}{r}\binom{m-1}{m-r}\tag{2}\\
&=m\sum_{r=1}^n\binom{n}{r}[x^{m-r}](1+x)^{m-1}\tag{3}\\
&=m[x^m](1+x)^{m-1}\sum_{r=1}^n\binom{n}{r}x^r\tag{4}\\
&=m[x^m](1+x)^{m-1}\left((1+x)^n-1\right)\tag{5}\\
&=m[x^m](1+x)^{n+m-1}\tag{6}\\
&=m\binom{n+m-1}{m}\tag{7}\\
&\,\,\color{blue}{=n\binom{n+m-1}{n}}\tag{8}
\end{align*}

Comment:


*

*In (2) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{p-q}$.

*In (3) we use the coefficient of operator according to (1).

*In (4) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

*In (5) we apply the binomial theorem.

*In (6) we multiply out and ignore the second term $1$ since it does not contribute to $[x^m]$.

*In (7) we select the coefficient of $x^m$.

*In (8) we use the binomial identity $\binom{p}{q}=\frac{p-q}{q}\binom{p}{q-1}$.
A: If generating functions are insisted upon I would write
$$\sum_{r=1}^n r {n\choose r} {m\choose r}
= \sum_{r=0}^n r {m\choose r} {n\choose n-r}.$$
Now what we have here is the Cauchy Product of two series
$$f(z) = z ((1+z)^m)' = m z (1+z)^{m-1}
\quad\text{and}\quad
g(z) = (1+z)^n.$$
The desired sum is the coefficient on $[z^n]$ of $f(z) g(z)$ or
$$[z^n] f(z) g(z) = [z^n] m z (1+z)^{n+m-1}.$$
The right is the generating function we wanted to find:
$$h(z) = f(z) g(z) = m z (1+z)^{n+m-1}.$$
We can extract the coefficient to get
$$ m [z^{n-1}] (1+z)^{n+m-1} = m {n+m-1\choose n-1}
= m\frac{(n+m-1)!}{(n-1)! \times m!}
\\ = n\frac{(n+m-1)!}{n! \times (m-1)!}
= n {n+m-1\choose n}.$$
Here we have used the fact that for formal power series
(this is the Cauchy Product)
$$[z^n] f(z) g(z) =
\sum_{r=0}^n [z^r] f(z) [z^{n-r}] g(z).$$
A: Hint : $\binom{n}{r}x^n=\frac{1}{r!} x^r \frac{d^r}{dx^r} x^{n}$
Using it you can find
\begin{align*}
\sum_{n=0}^{\infty}\sum_{r=1}^n r\binom{n}{r}\binom{m}{r}x^n &= \sum_{r=1}^n r \binom{m}{r} \sum_{n=r}^{\infty} \binom{n}{r} x^n\\
&= \sum_{r=1}^n r \binom{m}{r}\frac{1}{r!} x^r \frac{d^r}{dx^r} \sum_{n=r}^{\infty} x^{n}\\
&=\dots
\end{align*}
