Discrete Math problem with a deck of cards

Question

I have a standard deck of 52 cards. How do I find the number of hands of 13 cards that contain 4 cards of the same rank? (A,2-10, J,Q,K)

My initial thought was to choose 10 cards from 49, because 9 are normal cards and the 10th can be any random starting card that will determine what the next 3 cards are. The choose 3 cards from 3 as they must all be the same rank. This must be done for each rank, so it is brought to the 13th power.

$${49 \choose 10}{3 \choose 3}^{13} = {49 \choose 10}^{13}$$

I'm very confused with this question, is this on the right track?

Thanks

Answer 1

Your approach fails because you are assuming that the last three cards match the tenth. There are many more ways for the hand to have four of a kind.

The number of hands that include $4$ aces is ${48 \choose 9}$ because you have that many ways to choose the rest of the cards. It is tempting to multiply this by $13$ for the number of hands that include four of a kind, but you count hands with two four of a kinds twice, so you need the inclusion-exclusion principle.