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From a standard deck of 52 playing cards, find how many five-card hands can be dealt:

a) consisting of three twos and another pair

b) consisting of one pair and three of a kind

What I did: 4C3 * 22C1

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  • $\begingroup$ What is your rationale for $\binom{24}{1}$? $\endgroup$ – N. F. Taussig Dec 23 '16 at 0:30
  • $\begingroup$ Well, two pairs of each 3,4,5,6,7,8,9,10, Jack, Queen, King, Ace so thats selecting one pair from 22 $\endgroup$ – TripleA Dec 23 '16 at 0:32
  • $\begingroup$ How are you getting the number 22? $\endgroup$ – Barry Cipra Dec 23 '16 at 0:36
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From a standard deck of $52$ playing cards, find how many five-card hands can be dealt consisting of three twos and another pair.

There are $\binom{4}{3}$ ways of selecting three of the four twos. There are twelve ranks from which we can draw the other pair and $\binom{4}{2}$ choices for the cards of that rank. Thus, the number of hands that contain three twos and another pair is $$\binom{4}{3}\binom{12}{1}\binom{4}{2}$$

From a standard deck of $52$ playing cards, find how many five-card hands can be dealt consisting of one pair and three of a kind.

Hint: Choose one of the thirteen ranks for the pair, then choose two cards of that rank. Choose one the remaining twelve ranks for the three of a kind, then choose three cards of that rank. This hand is called a full house.

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There is a total of $\binom{4}{3}=4$ triplets of twos. Now, there is a total of $\binom{4}{2}=6$ pairs of each of the 12 remaining ranks, that amounts to $6*12=72$. Multiply that by 4 to get 288.

For the second question you can simply multiply the previous answer by 13 to get 3744.

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