TITLE: How many bit strings of length n contain exactly r 1s?
QUESTION [5 upvotes]: This has me confused. Wouldn't we want to use permutations in this case, and not combinations, since order is relevant when forming bit strings?

REPLY [3 votes]: Suppose $n=10, r=3$.  If you used permutations, then selecting the first, second, and third bits (to be 1 instead of 0) is one possible choice.  Selecting the third, second, and first bits is a different one.  But you don't want to count these differently, you want to count these the same.  Using combinations, selecting $\{1,2,3\}$ is one way of choosing three positions, and $\{1,3,8\}$ is another one.