Choose 13 cards from a standard deck of cards. How many 4-of-a-kinds are there?

Question

I believe the answer is 13 * $13\choose4$ * $48\choose9$.

There are $13\choose4$ to draw 4 of the same cards, and multiply by 13 for each possible rank (A, 2, 3, ..., K). Then there are $48\choose9$ to choose the remaining cards.

One thing I am not certain of, is whether this accounts for the possibility of having two 4-of-a-kinds or three 4-of-a-kinds, but I believe it is, since having two and three means you have one.

Answer 1

There is only 1 way to get four of a kind

13 * $4\choose4$ * $48\choose9$

Answer 2

There are $13$ ways to choose the rank of a four of a kind, one way to select all the cards of that rank, and $\binom{52 - 4}{13 - 4} = \binom{48}{9}$ ways to select the other nine cards in the hand from the other $48$ cards in the deck. Hence, there are $$\binom{13}{1}\binom{48}{9}$$ $13$-card hands that contain a four of a kind.

However, we have counted hands with two four of a kinds twice, once for each way we could designate one of the ranks as the one from which the four of a kind was drawn. We only want to count them once, so we must subtract them from the total.

There are $\binom{13}{2}$ ways to select two ranks from which four of a kinds are drawn, one way to draw all the cards of those ranks, and $\binom{52 - 2 \cdot 4}{13 - 2 \cdot 4} = \binom{44}{5}$ ways to choose the remaining five cards in the hand from the remaining $44$ cards in the deck. Hence, there are $$\binom{13}{2}\binom{44}{5}$$ $13$-card hands with two four of a kinds.

However, we have subtracted too much. We counted hands with three four of a kinds three times when we counted hands that contain a four of a kind, once for each way we could designate one of the ranks as the rank from which a four of a kind is drawn. We subtracted those hands three times when we counted hands that contain two four of a kinds, once for each $\binom{3}{2}$ ways we could designate two of those three ranks as the ranks from which two four of a kinds were drawn. Hence, we have not counted hands with three four of a kinds at all.

There are $\binom{13}{3}$ ways to choose the ranks of the three four of a kinds, one way to choose all the cards of each of those ranks, and $\binom{52 - 3 \cdot 4}{13 - 3 \cdot 4} = \binom{40}{1}$ ways to choose the remaining card in the hand from the $40$ remaining cards in the deck. Hence, there are $$\binom{13}{3}\binom{40}{1}$$ hands with three four of a kinds.

By the Inclusion-Exclusion Principle, the number of $13$-card hands that contain a four of a kind is $$\binom{13}{1}\binom{48}{9} - \binom{13}{2}\binom{44}{5} + \binom{13}{3}\binom{40}{1}$$