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You are dealt $5$ cards from a standard deck of 52 cards. A standard deck of 52 cards has 13 ranks in each of four suits.) How many ways can you be dealt the $5$ cards so that they contain two cards of one rank, two cards of another rank, and a fifth card of a third rank? We say that such a hand has two pairs. For example, the hand QQ225 has two pairs. (Assume that the order of the cards does not matter.)

I'm not sure how to start.. Seems like permutations, but I'm not sure. Any help would be appreciated, and thank you in advance.

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    $\begingroup$ I would avoid using the word "permutations" for anything except very explicitly arrangements of unique items such that all items are used exactly once each and order matters or for the algebraic object referring to a bijective function from a finite set to itself. These are "combinations" instead. $\endgroup$
    – JMoravitz
    Commented Sep 21, 2020 at 19:29
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    $\begingroup$ As for how to start... understand binomial coefficients. Understand rule of product. From there, approach like you would any other rule of product question. Pick the ranks that are used for the pairs. Pick the suits used for the smaller of the pairs. Pick the suits used for the larger of the pairs. Finally, pick the remaining card. $\endgroup$
    – JMoravitz
    Commented Sep 21, 2020 at 19:30

1 Answer 1

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$$\binom{13}{2}\times \binom{4}{2}\times \binom{4}{2}\times 44=123,552$$

  1. $\binom{13}{2}$: choose the two pairs

  2. $\binom{4}{2}$: choose the suits of the first pair

  3. $\binom{4}{2}$: choose the suits of the second pair

  4. $52-8=44$: any of the remaining cards not ranked with the two pairs

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