Just a simple poker question using combinatorics.
Consider a standard deck of 52 cards. How many different four card hands have two pairs?
First off, this is generally not an elementary level question. But getting down to business: How many different combinations of card numbers are there for the pairs? For instance, one possibility is K, 8. There are 13 different card "numbers", so the answer is $\displaystyle _{13}C_2 = 78$. Now, of each of those, how many ways can we choose distinct suits for the cards? Well, there are six ways for each pair, so that gives us 36 total. So the answer is $\displaystyle 78 \cdot 36 = 2808$.