I'm getting tripped up because of the "exactly" part. Can anyone explain how to approach this problem and ones that may be similar. For example, if I wanted a hand that contained exactly three K's, or one K and three 9's.
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asked Nov 14, 2012 at 1:30
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$\begingroup$ One thing that may be helpful if you are stuck is to solve a simpler problem where you can enumerate things to make sure that you have your reasoning right. For example, you could ask yourself how many 3-card poker hands with 1 ace can you make from a deck of cards that has 2 aces and 5 non-aces. Varying the small numbers can also help you find patterns that might allow you to guess the form of the answer. $\endgroup$– Michael JoyceNov 14, 2012 at 3:41
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1 Answer
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There are four ways to pick an ace. The remaining four cards cannot be an ace, and so there are $\binom{48}{4}$ ways to choose them.
Answer. $4\binom{48}{4}$.
