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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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. $4\binom{48}{4}$. |
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