Is it possible to prove $\sum_{k=0}^n \binom{n\vphantom{k}}{k} \binom{k}{m} = \binom{n\vphantom{k}}{m} 2^{n-m}$ combinatorially? $$\sum_{k=0}^n \binom{n}{k} \binom{k}{m} = \binom{n}{m} 2^{n-m}$$
So for the proof, I have to use a real example, such as choosing committees, binary sequences, giving fruit to kids, etc. I have been thinking hard on it, but I don't fully understand it to do so.  Any help is really appreciated. 
 A: Start with an $n$-element set $N$, say $\{1,2,\dots,n\}$. The $k$-th term in the sum on the left side of your equation counts the number of ways of choosing, first, a $k$-element subset $K$ of $N$, and then an $m$-element subset $M$ of $K$.  So the whole sum counts the ways of making such choices, $M$ and $K$ such that $M\subseteq K\subseteq N$, if you allow $K$ to be of any size (but $M$ still has to have $m$ elements).  The right side of your equation counts the same thing, because there are $\binom nm$ possibilities for $M$ and, once you've chosen $M$, choosing $K$ amounts to choosing a subset of the $(n-m)$-element set $N-M$ to serve as the rest of $K$ (meaning to serve as $K-M$).
A: Say you have n people and a large empty ship. There are needed exactly m people (crew) to operate the ship, and there could be any number of passengers. In how many ways can you select the crew and the passengers from the group of n people?
First way: Select the crew first in $\binom{n}{m}$ ways. For every one of the remaining $n-m$ people you have 2 choices: either to select him as a passenger or not. So overall you have $\binom{n}{m}2^{n-m}$ ways of filling the ship.
Second way: Select first which people will get on the ship (not making distinction between crew and passengers yet). Then select which of them will be the crew. If you had made your mind that $k$ people will get on the ship then you could select them in $\binom{n}{k}$ ways, and then select the crew members between them in $\binom{k}{m}$ ways. Since $k$ could be any number from $m$ to $n$ you should take the sum over $k$ of the above product. Starting the summation from a smaller value of $k$ is ok since $\binom{k}{m}=0$ for $k<m$.
