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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?

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  • $\begingroup$ The problem you are solving in the body seems to be different from the one stated in the title. $\endgroup$ – N. F. Taussig Mar 8 '17 at 20:02
  • $\begingroup$ @N.F.Taussig I edited my OP. $\endgroup$ – vasya pupkin Mar 8 '17 at 20:04
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You correctly calculated the number of ways of selecting an Ace and a club without choosing the Ace of clubs since we must choose one of the other three Aces and one of the other twelve clubs.

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 $$\binom{3}{1}\binom{12}{1}\binom{36}{3}$$

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