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Math Help - Bridge hands

  1. #1
    Member oldguynewstudent's Avatar
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    Bridge hands

    Nate, Ben, Suzy, and Gracie play bridge. In how many ways can the 52-card deck be dealt so that each player receives 13 cards?

    Since all cards are being dealt, wouldn't this just be \left(52\right)_{52}?

    My reasoning is that in the 52! permutations of the cards each player can just get every fourth card, so it doesn't matter about the 13 card hands. It only matters in how many ways all 52 cards can be selected from the deck.
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    Nate, Ben, Suzy, and Gracie play bridge. In how many ways can the 52-card deck be dealt so that each player receives 13 cards?

    Since all cards are being dealt, wouldn't this just be \left(52\right)_{52}?

    My reasoning is that in the 52! permutations of the cards each player can just get every fourth card, so it doesn't matter about the 13 card hands. It only matters in how many ways all 52 cards can be selected from the deck.
    Sorry I am unfamiliar with the notation \left(52\right)_{52}.

    I think the answer can be expressed as \binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{1  3}{13}. Choose the first player's cards, then the second player's, etc.
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  3. #3
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    Hello, oldguynewstudent!

    Nate, Ben, Suzy, and Gracie play bridge.
    In how many ways can the 52-card deck be dealt so that each player receives 13 cards?

    undefined is absolutely correct!


    Another way to write it is: . {52\choose13,13,13,13}


    This is a partition of 52 objects into 4 ordered sets of 13 objects.

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    Just to expand a bit on the answers you have already been given, it all depends on what is meant by "the number of ways".

    If the order of the cards in each hand matters, your original answer (52!) is correct. But usually we think the order is not significant, in which case the correct answer is that given by undefined and Soroban.
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