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Math Help - From a 52 card deck, how many 4 card hands ...

  1. #1
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    From a 52 card deck, how many 4 card hands ...

    From a 52 card deck, how many 4 card hands are there that

    (a) Do not contain a pair? (That is, the ranks are all different)
    (b) Have exactly 2 pairs?

    Any help would be great! Thanks!
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  2. #2
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    Quote Originally Posted by jlt1209 View Post
    From a 52 card deck, how many 4 card hands are there that
    (a) Do not contain a pair? (That is, the ranks are all different)
    (b) Have exactly 2 pairs?
    Why don't you tell us what you know about this problem?
    Just giving you a complete solution will not help you understand anything about this subject.
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  3. #3
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    Well I'm pretty sure the answer to part a is 52*48*44*40/4! because you can pick the first card in 52 ways, then if the second isnt a pair you have to avoid 3 cards so its 48, and so on. And then divide by 4! because order doesn't matter.

    The second part I'm having more trouble with. I think it is something like 52*3*48*3/4! but I don't think that's exactly right.
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  4. #4
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    Hello, jlt1209!

    From a 52 card deck, how many 4-card hands are there that

    (a) Do not contain a pair? (That is, the ranks are all different)
    Your answer is correct!


    (b) Have exactly 2 pairs?
    First, select the ranks of the two pairs.
    . . There are: . {13\choose2}  \:=\:78 choices.

    Then select the two pairs.
    . . For the first pair, there are: . {4\choose2} \:=\:6 ways.
    . . For the second pair, there are: . {4\choose2} \:=\:6 ways.

    Therefore, there are: . 78\cdot6\cdot6 \:=\:2808 posible two-pair hands.

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  5. #5
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    Ohh ok that makes sense. Thanks so much!
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  6. #6
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    Why can it not be the OP's answer of (52*3)(48*3) though?

    I don't see why you cannot first a pick a card out of the 52, then have only 3 cards to pick from to make a pair. And then, out of the 48 remaining, pick another one out of 3 to make the second pair.

    It's confusing to me why this is wrong.
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