# Combination of cards

From a deck of 52 cards, how many five card poker hands can be formed if there is a pair (two of the cards are the same number, and none of the other cards are the same number)?

I believe you can pick out the first card by ${_4}C_2$, as there are 4 cards which would be the same number (as there are 4 suits). I would pick two from here.

From here on though, I am unsure. I believe it involves the numbers 48, 44, and 40, as after every picking you cannot have an identical number anywhere, so there would be 4 less to choose from. However, I don't believe I can just do $_{48}C_1 * _{44}C_1,..$ as I am not simply selecting 1 card from 48 and removing 4 random ones.

The answer is $1098240$.

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We will use the notation $\binom{n}{r}$, which is more common among mathematicians, where you write ${}_nC_r$.
The kind of card that we have a pair of can be chosen in $\binom{13}{1}$ ways. For each choice of kind, the actual cards can be chosen in $\binom{4}{2}$ ways. (By kind we mean things like Ace, or $7$, or Queen.)
For each choice made so far, we now count the number of ways to pick the rest of the cards. The kinds of cards we have singletons of can be chosen in $\binom{12}{3}$ ways. For each such choice, the actual cards can be chosen in $4^3$ ways. This is because if, for example, we are to have a $7$, a $10$, and a Queen, the $7$ can be picked in $4$ ways, as can the $10$, as can the Queen. The total number of one pair hands is therefore $$(13)\binom{4}{2}\binom{12}{3}4^3.$$ Compute. We get $1098240$.
 Thank you. I'm just going to reiterate what I believe you mean here. The 13C1 ways are from Ace to King. From that, there are 4 suits from which we choose 2 of to get the pair. Now there are 12 numbers (say Ace to Queen) we can choose from. We choose 3 of these (2 have been chosen). For each of these three, there are 4 suits again. By the fundamental counting principle, there are then 4*4*4 ways to choose it (4 for each as there are 4 suits). Is that correct? – InQ Jun 10 '12 at 19:23 Yes, good summary, shorter than mine. The only thing I don't understand is your sentence "We choose $3$ of these ($2$ have been chosen)." The part in parentheses seems not relevant. Two cards have already been chosen, of the same kind, but that has nothing to do with choosing $3$ kinds of cards to have singletons in. – André Nicolas Jun 10 '12 at 19:30 Yes, sorry, that's what I meant. Two cards have been chosen, three more cards to choose from, and to determine that, choose three numbers (kinds) that each have 4 suits. By 'have singletons in', you simply mean just to have one of each kind of card, right? – InQ Jun 10 '12 at 19:35 Yes, I mean one of that kind. – André Nicolas Jun 10 '12 at 19:37 Thank you for all your help! The connections to the problem was easy to make thanks to your answer. – InQ Jun 10 '12 at 19:40