TITLE: A simple "number of bit strings", so I thought
QUESTION [7 upvotes]: How many bit strings of length 10 contain at least three 1s and at least three 0s?

REPLY [3 votes]: Consider the complementary problem: at most 2 0s or at most 2 1s. The counts for each case is $\binom{10}{2}+\binom{10}{1}+\binom{10}{0}=56$; the count for at least 3 0s and at least 3 1s is therefore $2^{10}-2\cdot 56=912$