Combinatorial proof of: $$\sum_{k=0}^n \binom{n}k \binom{n-k}{m-k} = 2^m \binom{n}m$$
My answer is:
number of ways to form 2 committees of different members from a group of n people.
For the first method, choose k members of the first committee from n people, and form the other committee with (m-k) members from the remaining (n-k) people, since k members are already in the first committee. The product is the number of ways.
For the second method, choose m members who will be part of the two committees from n people. The number of ways to form committees from m people is the subset of m members, which is $2^m$. The product is the number of ways.
Did I get some things right?
$\newcommand{\Set}[1]{\left\{ #1 \right\}}$You are nearly there. You have $n$ people. You want to select $m$ of them, and then divide these $m$ people in a group labelled $0$, and a group labelled $1$.
On the right-hand side, you choose a subset $A$ of $m$ people in $\dbinom{n}{m}$ ways, and then divide these $m$ people in the two groups, by choosing one of the $2^{m}$ characteristic functions $A \to \Set{0, 1}$, which tells you whether person $a \in A$ goes into group $0$ or $1$.
On the left-hand side, for every $k \le m$ you first choose $k$ people to go in the group labelled $0$, and then $m - k$ people to go in the group labelled $1$.