# Card Hand Combinations

A probability question asks how many ways one can select a 5 card one pair hand (that is a pair of cards are the same number and the other 3 are different) from a standard 52 card deck.

The answer for the question from the back of my book was calculated as $$\frac{13\pmatrix{4\cr2}\cdot12\pmatrix{4\cr1}\cdot11\pmatrix{4\cr1}\cdot10\pmatrix{4\cr1}}{3!}$$ Why is the denominator not $4!$ since there are 4 groups that we are trying to arrange?

How is this any different than if you have 5 letters a, b, c, d, e and you are trying to find the amount of ways you can get groups of 3 using $$\frac{5\cdot4\cdot3}{3!}$$ since there are 3 elements to be arranged so you need to divide out the 3! permutations an ordered triplet can produce?

Thanks

EDIT: Second example was worded wrong, corrected now

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Example: if your hand has two 10s, and one each of 8, 7, and 6, then that's the same as if your hand has two 10s, and one each of 6, 7, and 8. It is not, however, he same as if your hand has two 8s, and one each of 10, 7, and 6. In other words, the number you have two of cannot be permuted with the other numbers. –  Will Orrick Apr 20 '12 at 23:13

I would solve the problem in a marginally different way. The kind you have $2$ of can be chosen in $\binom{13}{1}$ ways. For each of these ways, the actual cards can be chosen in $\binom{4}{2}$ ways. So far the same as yours. Then the kinds you have $1$ each of can be chosen in $\binom{12}{3}$ ways. Once you have chosen these kinds, line them up in order of strength. In that order, choose the actual cards. This can be done in $\binom{4}{1}^3$ ways, for a total of $$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3.$$ This is the same answer as yours, with the division by $3!$ "hidden" in $\binom{12}{3}$.
If, as you did, you use the product $\binom{12}{1}\binom{11}{1}\binom{10}{1}$, then every one of the $3$ useless card choices is counted a total of $3!$ ways. So we divide by $3!$. We do not divide by $4!$ because it is only the singleton cards that have been multiply counted. What has happened is a kind of hybrid counting, some without order (what we have $2$ of) and the rest in order.
We could count in a way that forces division by $5!$ (note, not $4!$) by following the order of the dealing. Choose the kind we have $2$ of. This can be done in $\binom{13}{1}$ ways. Now choose when these cards come to you. Their locations in the dealing can be chosen in $\binom{5}{2}$ ways. Now fill the leftmost such location with a card of the right kind. There are $\binom{4}{1}$ ways to do this. Then fill the second such location. this can be done in $\binom{3}{1}$ ways. Now fill the empty locations, from left to right. This can be done in $\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}\binom{10}{1}\binom{4}{1}$ ways, for a total of $$\binom{13}{1}\binom{5}{2}\binom{4}{1}\binom{3}{1}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}\binom{10}{1}\binom{4}{1}.$$ We have counted every hand $5!$ times, so finally we need to divide by $5!$.
If you want to think of it that way, fine. But since there is only one such weird card, it didn't get multiply counted. But consider the "two pairs" poker problem. One approach is $\binom{13}{1}\binom{4}{2}\binom{12}{1}\binom{4}{2}(44)$. That double counts, need to divide by $2!$. For not double counting, use $\binom{13}{2}\binom{4}{2}\binom{4}{2}(48)$. The $\binom{13}{2}$ is from choosing the kinds we have two each of. –  André Nicolas Apr 20 '12 at 23:12