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Math Help - Combination with pack of cards

  1. #1
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    Combination with pack of cards

    I am reading over some of my notes and I am having a little trouble understanding combinations and permutations and when to use them.

    Problem: How many thirteen-card hands can be chosen from a pack of cards?

    My Solution:The order of cards once in the hand does not matter so am I right by saying there are P(52,13)/P(13,13) ways. I.e. C(52, 13) ways?

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    Problem:On the hands in the previous problem, how many consist of seven cards in one suit and six in another?

    Sample Solution:

    1. Pick a suit..... 4 ways
    2. Choose seven cards from this suit, with repetition forbidden and order not important .... C(13,7) ways
    3. Pick another suit .... 3 ways
    4. Choose six cards from this suit ... C(13,6) ways

    So number of hands is 4*C(13,7)*3*C(13,6) => P(4,2)C(13,7)C(13,6)

    This is the solution in my notes but I don't understand in step for why there is C(13,6) ways. Shouldn't there only be C(6,6) ways since we have already picked the first 7 cards?

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    Problem: What is the number of 12 card hands containing four cards in each of three suits?

    Sample Solution: P(4,3)C(12,4)C(12,4)C(12,4) = P(4,3) C(12,4)^3

    Again, with this solution why isn't it P(4,3)C(12,4)C(8,4)C(4,4)

    Also shouldn't it be C(4,3) not P(4,3) since the order we pick the suits doesn't matter. Ie. 4 Clubs, 4 Spades, 4 Diamonds is the same as 4 Diamonds, 4 Spades, 4 Clubs.

    Help would be much appreciated. D:
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  2. #2
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    Re: Combination with pack of cards

    C(52,13) is correct. In your 2nd problem its not C(6,6) because they are selecting 6 from a different suit.
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  3. #3
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    Re: Combination with pack of cards

    Take a look at Durango Bill's web site.

    Durango Bill's Home Page
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  4. #4
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    Re: Combination with pack of cards

    Scroll dowm to Bridge Combinatorics and Probabilities and then click on (the various) statistics.
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  5. #5
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    Re: Combination with pack of cards

    Quote Originally Posted by biffboy View Post
    C(52,13) is correct. In your 2nd problem its not C(6,6) because they are selecting 6 from a different suit.
    I'm still unsure. C(13,7) means we are choosing 7 cards from 13 possibilities where order doesn't matter and repetitions aren't allowed. After we have picked our 7 cards we only have 6 cards remaining so why don't we use C(6,6)? I am sure what you are saying is right I just don't understand the concept behind it, I think I am missing some key thing about how combinations work.

    Quote Originally Posted by BobP View Post
    Scroll dowm to Bridge Combinatorics and Probabilities and then click on (the various) statistics.
    Thanks. The website helped me a little. Especially looking at poker probability, however I am still uncertain of the third problem I pose.
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  6. #6
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    Re: Combination with pack of cards

    For your second problem there are twelve possible suit combinations; 4 suits to choose from for the 7 card suit and then 3 suits to choose from for the 6 card suit. 4*3=12
    For the 7 card holding, there are13C7 possible holdings. (C(13,7) in your notation.)
    For the 6 card holding, there are still 13 cards to choose from, you haven't taken any cards from this suit yet. So, 13C6 possibilities.
    (BTW stick to Combinations throughout in these calculations rather than Permutations. Only if you are actually bothered about the order that things happen do you need to worry about Permutations.)
    The total number of possible hands is therefore 4*3*13C7*13*C6.

    For your third problem, the number of 4 card holdings in a particular suit is 13C4. If the three suits are, for example, spades, hearts and diamonds, then you can have any 4 of the 13 spades with any 4 of the 13 hearts with any 4 of the 13 diamonds. So if these happen to be the three suits the number of possible holdings will be 13C4*13C4*13C4. There are four possible three card suit combinations so you multiply by 4.
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