# Poker hands with one Ace and one club without Ace of clubs

First we choose $2$ cards. Start by picking a rank for Ace. Can be done in $\binom 11$ ways. Then we pick a suit for it: $\binom 11 \binom 31.$ After that we choose the second card. Pick a rank from $12$ remaining and choose a suit for it: $\binom {12}1 \binom11$ because there's only one way to choose a club out of $4$ suits. So far we have $\binom 11 \binom 31\binom {12}1 \binom11$ combinations for the first two cards.

Since the remaining three cards can have the same rank we can choose them in $\binom{12}1^3$ ways. Because of the possibly repeating ranks we must choose suits in nonrepeating way: $\binom{12}1^3 \binom31 \binom21 \binom11.$ But this overcounts the number of the last three cards. What's wrong with counting the last three cards this way?

• The problem you are solving in the body seems to be different from the one stated in the title. – N. F. Taussig Mar 8 '17 at 20:02
• @N.F.Taussig I edited my OP. – vasya pupkin Mar 8 '17 at 20:04

Once the Ace and the club have been selected, we are left with $52 - 3 - 12 = 36$ cards since we cannot use one of the other three Aces or one of the other twelve clubs. To obtain a full poker hand, we must choose three of those $36$ cards. Therefore, the number of poker hands that contain exactly one Ace and exactly one club that do not contain the Ace of clubs is $$\binom{3}{1}\binom{12}{1}\binom{36}{3}$$