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How many ways can you choose at least one A from a deck of card in a poker hand?

I just wanted to double check my answer, would it be

C(52,5)- C(48,5)

Help is much appreciated,

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  • $\begingroup$ Yes, your calculation gives the number of hands with at least one Ace. $\endgroup$ – André Nicolas Apr 23 '14 at 19:32
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To get at least one ace is to get 1, 2, 3, or 4. You are selecting the aces among the four aces, the other cards among the $52 - 4 = 48$ non-aces. In all: $$ \binom{4}{1} \cdot \binom{48}{4} + \binom{4}{2} \cdot \binom{48}{3} + \binom{4}{3} \cdot \binom{48}{2} + \binom{4}{4} \cdot \binom{48}{1} $$ Or you could say there are $\binom{52}{5}$ hands in all, of those $\binom{48}{5}$ are ace-less, which gives: $$ \binom{52}{5} - \binom{48}{5} $$

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