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Suppose 3 people are each dealt 2 cards from one standard deck of playing cards. You may assume that the order in which the cards are dealt does not matter. In how many ways can the cards be dealt so that everyone gets at least one ace?

So there are 4 different ways to pick the first ace, then 3 for the second, and 2 for the third.

So $4\times3\times2$

Then $\binom{49}{3}$ ways to pick 3 more cards to distribute, and 3! ways to distribute them.

So I get $4(3)(2)\binom{49}{3}3! = 2653056$ but the answer I'm given is $2571840$ so I'm clearly over counting somehow, I'm guessing it has to do with how the last ace is distributed? How do I handle that?

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Case I, nobody gets the fourth ace. The number of distributions is $$4\times3\times2\times48\times47\times46\ .$$

Case II, all four aces are received. Begin by choosing who gets two aces, and which two he gets. The number of distributions is $$3\times6\times2\times48\times47\ .$$

The total is $$3\times2\times48\times47\times(4\times46+6)=2571840\ .$$

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