In how many ways can you deal a "full house"? (Three cards of one rank and two cards of another rank.)
What is wrong with this approach?
${13 \choose 2}$ ways to choose 2 ranks out of 13
${4 \choose 2}$ ways to choose 2 suits out of 4
${4 \choose 3}$ ways to choose 3 suits out of 4
${13 \choose 2}{4 \choose 2}{4 \choose 3}$
The error is in your first term. When you use $\binom{13}{2}$, you're saying that the order doesn't matter: For example, we should treat a 7 and a queen the same as a queen and a 7. However, when you multiply by the other terms, you need to pick which one your values gets three suits and which one gets two. For example, two sevens and three queens is different from two queens and three sevens.
So we have to sort of take order into account, at least to the extent that we select a value and then select which two suits it gets, then select a second value and then select which three suits it gets. So there are $\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{3}$ possible full houses.