How to derive this binomial identity? I believe the following is an identity (I've tested with a few random $m$ and $n$ values, could be wrong though):
$$\sum_{k= 0}^{\infty}{m \choose k}{n \choose k}k=n\binom{m+n-1}{m-1}$$
but I'm not sure how to prove it. 
Initially I thought of considering the right side combinatorially: pick $m-1$ things from a bucket of $m+n-1$ things and repeat this $n$ times. But how do I think of the left side combinatorially, since there is an infinite sum? Or maybe this is the incorrect approach?
 A: Hint
Note that $k\binom{n}{k}=n\binom{n-1}{k-1}$, and erase $n$ from both sides. 
There is one box with $m$ balls, and the other with $n-1$ balls. 
You are trying to pick $n$ balls in total.
Note that this is equal to $(RHS)$.

 Note that $\binom{n-1}{k-1}=\binom{n-1}{n-k}$
 Since you are trying to pick $n$ balls in total if you pick $k$ balls from the first box, you are going to have to pick $n-k$ from the second. This operation is equal to $(LHS)$. 

A: $$\begin{align}
\sum_{k= 0}^{\infty}\color{green}{{m \choose k}}\color{blue}{{n \choose k}k}
&=\color{blue}n\sum_{k=0}^\infty\color{green}{\binom m{m-k}}\color{blue}{\binom {n-1}{k-1}}\\
&=n\binom{m+n-1}{m-1}\qquad\blacksquare\end{align}$$
using the Vandermonde identity. Applicable limits for $k$ are $1\le k\le \min(m,n)$ as $\binom ab=0$ for $a<b$.
A: Since you were originally aiming for a combinatorial proof, I thought I'd offer one: a committee consists of $m$ members from the minority party and $n$ members from the majority party.  A subcommittee is to be formed having an equal, but unspecified, number of members from each party, and one of the subcommittee members from the majority party is to be designated chair of the subcommittee.  Call the result a marked subcommittee.
The left side clearly enumerates the number of marked subcommittees.  To see that the right side does too, observe that the majority-party and minority-party members of a marked subcommittee may be paired up as follows: line up the subcommittee members from the majority party with the chair first and everybody else in alphabetical order.  Line up the subcommittee members from the minority party in alphabetical order.  Now pair up the first people in each of the lineups, the second people in each of the lineups, and so on.
We can now interpret the right side of the identity: select a person from the majority party to be subcommittee chair.  Now line up the $n$ majority-party members with the chair first and everybody else in alphabetical order.  There $n$ possible lineups.  Place $m-1$ separator marks in this lineup to partition the lineup into $m$ parts (some possibly empty).  By stars-and-bars, there are $n\binom{n+m-1}{m-1}$ partitioned lineups.  Line up the minority-party members in alphabetical order and pair up the first minority-party member with the first part of the majority-party lineup, the second minority-party member with the second part of the majority-party lineup, and so on.  Then the subcommittee is formed by taking those minority-party members that have been paired with a non-empty part of the majority-party lineup, and by taking as majority-party members the lead member of each such non-empty part.  Note that since the chair will be the lead member of her/his part, the chair will always be taken.
Example: To show how the correspondence works, let the majority party consist of $A$, $B$, $C$, $D$, $E$ and the minority party of $1$, $2$, $3$, $4$.  Suppose the subcommittee consists of $B$, $D$, $2$, $3$, with $D$ as chair.  Then the pairing between minority-party and majority-party subcommittee members is $(2,D)$, $(3,B)$.  The corresponding partition of the majority-party lineup is $\lvert DA\lvert BCE\lvert$, the associated pairing is $(1,())$, $(2,(D,A))$, $(3,(B,C,E))$, $(4,())$.
Going the other way, suppose $B$ is chosen as chair and the partition of the majority-party lineup is $B\lvert\lvert ACD\lvert E$.  The associated pairing is $(1,(B))$, $(2,())$, $(3,(A,C,D))$, $(4,(E))$, which produces the subcommittee consisting of $B$, $A$, $E$, $1$, $3$, $4$, with $B$ as chair.
