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A standard deck of cards contains $52$ cards divided into four suits: the red suits, hearts and diamonds, and the black suits, clubs and spades. Each suit, in turn, is divided in 13 ordered ranks: ace through 10 followed by the face cards jack, queen, and king. The ace can act as either the lowest or highest card in the ranking.

What is the number of $5$-card hands in a $52$-card deck that contain two pairs (i.e., two pairs from different ranks, and a fifth card of a third rank)?

Answer: $\binom {13}{2}{\binom 42}^2 \binom {44}{1}$

Why is $\binom {13}{2}$ a factor in the answer?

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First you choose the two different ranks (one for each pair) so you get :

$$\begin{pmatrix}13\\2\end{pmatrix}\text{ choices.} $$

For each one of these choices you must choose two different suits for the first rank :

$$\begin{pmatrix}4\\2\end{pmatrix}\text{ choices.} $$

you must choose two different suits for the second rank :

$$\begin{pmatrix}4\\2\end{pmatrix}\text{ choices.} $$

And you must choose a last card which is among the remaining ranks :

$$(13-2)\times 4=44\text{ choices. } $$

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  • $\begingroup$ Is it the more general version of this rule: Consider $2-$lists for which there are $n$ choices for the first element and for each choice of the first element there are m choices for the second element. Then the number of such lists is nm. For example one possible list is (Jack, Spade, 10). Is that correct? $\endgroup$ – keys Apr 27 '15 at 16:49

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