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What is the total number of ways in which the poker hand is full of house that is you have to pick 5 cards out of 52 cards such that it contains exactly 3 cards with the same value.

Example a card consisting of 3 sixes and 1 king and 1 queen is a full of house.

And my solution is for the first 2 card can be different and rest of remaining cards needed to be same

13*12*11 = 1716

I think my approach is wrong as the answer given in the book is 3744.

Can someone tell me what I am missing and how can I extend this to pick 5 card such that atleast 3 of them have the same value.

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  • $\begingroup$ How did you come up with 13 * 12 * 11? $\endgroup$ – Christopher Liu Nov 22 '14 at 17:59
  • $\begingroup$ I need to fill 5 blanks lets fill the first with any of 13 and 2nd with any of 12 now I am left with 11 all remaining needed to be same. $\endgroup$ – akashchandrakar Nov 22 '14 at 18:04
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What you are describing is not a full house, but a three of a kind. For a full house, there will be $3$ cards of the same value, and the remaining $2$ cards will share a different value. For example, three sixes and two queens is a full house.

Don't forget that for each value, there is one card of each suit, for a total of $4$ cards of each value.

Now, to get a full house, you will need to pick two different values: one for the trio and one for the pair. We can do this in $13\cdot12$ different ways. Once we have chosen the value for the trio, we must pick three of the four suits, which we can do in $\binom43=4$ ways. Once we have chosen the value for the pair, we must pick two of the four suits, which we can do in $\binom42=6$ ways. Hence, we have $$13\cdot12\cdot4\cdot6=3744$$ different ways.

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I'm not quite sure your book was right. We can choose the hands with the following method: From 13 ranks choose 1 rank to be your three of a kind From that 1 rank, choose 3 out of the 4 suits From the remaining 12 ranks, choose 2 to get the ranks of the other two cards, which must be different For each card, there are 4 different ways to choose the suit. This gives us a total of $$\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}\binom{4}{1} = 54, 912$$. To extend this, think about where in this process we restricted the values of the remaining two cards.

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    $\begingroup$ The discrepancy is that the book computed the possibilities of a full house and the problem statement described a three-of-a-kind. The numbers are different, of course. $\endgroup$ – David K Nov 22 '14 at 19:06

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