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Math Help - How many 7-card hands can be chosen

  1. #1
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    How many 7-card hands can be chosen

    How many 7-card hands can be chosen from a standard deck of 52 cards such that no two cards are of the same rank?

    Is it as simple as  {13\choose 7} or am I missing something? I tried starting with all {52 \choose 7} possibilities and then subtracting the set of hands where there was exactly one pair, then exactly 3 the same etc. but it became quite convoluted.

    It seems like I am missing a lot of potential hands with  {13\choose 7} as there are four valid choices from each of the 13 ranks each time you choose - not just one.

    Would it be:

     {13\choose 1}\times 4 \times {12\choose 1}\times 4 ... \times {7\choose 1}\times 4 = \frac{13!}{6!}\times 4^7 ? but then what would I divide by to remove equivalent permutations?
    Last edited by terrorsquid; October 16th 2011 at 05:04 AM.
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  2. #2
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    Re: How many 7-card hands can be chosen

    Quote Originally Posted by terrorsquid View Post
    Would it be:

     {13\choose 1}\times 4 \times {12\choose 1}\times 4 ... \times {7\choose 1}\times 4 = \frac{13!}{6!}\times 4^7 ? but then what would I divide by to remove equivalent permutations?
    I think you divide by 7! because the order of cards does not matter. So, this is \binom{13}{7}\cdot4^7: first you select the rank and then for each rank you select the suit.
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