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Math Help - Counting Unordered Selections: Card Selection

  1. #1
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    Counting Unordered Selections: Card Selection

    From a standard deck of 52 playing cards, find how many five - card hands can be dealt:
    e)consisting of three twos and another pair

    Here is my attempt:
    3 Twos: 4C3
    Selection of Another Card with pair:
    48C1 x 3C1, acquiring

    576

    Solution however has 288 as answer. Someone point out flaw in logic.
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  2. #2
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    Note that your answer is exactly twice the correct answer. This is because you consider the remaining pair to be ordered. E.g., you first select a three of hearts and then a three of diamonds; but you could select them in the opposite order. These selections are counted separately in your approach.

    Another way is first to select the rank (3 - 10, J, Q, K, A) and then select an unordered pair.
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    So you are saying that 48C1 x 3C1 are inessence one selection? And that 4C3 and 4C1 x 3C1 are distinct? And could yous pecific other method of doing it?
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    So you are saying that 48C1 x 3C1 are inessence one selection?
    I am not sure what your question means. I am saying that 48C1 x 3C1 is the number of ordered pairs of cards of equal rank whose rank is from 3 to ace. Usually when a hand of cards is considered, the order of cards does not matter, so the question (in part) asks for the number of unordered pairs. This number is half the number above.

    And that 4C3 and 4C1 x 3C1 are distinct?
    Yes, 4C3 = 4, 4C1 x 3C1 = 4 x 3 = 12, so 4C3 and 4C1 x 3C1 are distinct numbers. Sorry to be a "Math Nazi": I understand that this is most likely not what your question means, but I don't want to spend too much time figuring out what it might mean.

    And could yous pecific other method of doing it?
    Sorry again, I did not even understand the grammatical structure of your sentence.
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    Quote Originally Posted by Lukybear View Post
    From a standard deck of 52 playing cards, find how many five - card hands can be dealt:
    e)consisting of three twos and another pair

    Here is my attempt:
    3 Twos: 4C3
    Selection of Another Card with pair:
    48C1 x 3C1, acquiring

    576

    Solution however has 288 as answer. Someone point out flaw in logic.
    Deck of cards is a set. One five-card hand is a subset. Subset implies combinations. First you choose three twos, this can be done in 4C3 number of ways. For each choice of twos you need to pick another pair. You have 12 x 4 cards you have to choose a pair from, so first you decide which pair should it be, and that you can do in 12C1 different ways. And last you have to choose which two card out of 4 chosen will you pick to form a pair, giving you the 4C2 possibilities. All in all, you ended up with 4C3 * 12C1 * 4C2=288.
    Last edited by MathoMan; September 14th 2010 at 01:39 AM. Reason: fixed grammatical errors
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    Thxs to above replies. To Matho, so how is one able to differentiate between dividing to remove repitition and not dividing such as here 4C3 * 12C1 * 4C2
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    You asked if someone could find what went wrong with your reasoning and emakarov did that.
    The difference is that your logic includes ordering of the last two cards while mine does not: you picked last two cards one by one 48C1 and 3C1 which implies that it matters which of two cards in the pair is the first one and which one is the second one, When it actually does not matter, while I picked them both at once 4C2.
    Last edited by MathoMan; September 14th 2010 at 03:22 PM.
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  8. #8
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    how is one able to differentiate between dividing to remove repitition and not dividing such as here 4C3 * 12C1 * 4C2
    When one uses a product rule to count the number of consecutive choices (e.g., first choose any of the 48 cards and then choose one of the three of the same rank), one is counting ordered pairs (or triples, etc.). In this case, one can choose card c1 and then another card c2, and this would be different from choosing c2 first and then c1. Therefore, the unordered pair (c1, c2) is counted twice. On the other hand, 4C2 is by definition the number of unordered pairs of cards of the same rank. Therefore, choosing a rank first and then choosing an unordered pair of this rank produces 12C1 x 4C2 possibilities.
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