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Math Help - Counting (Permutations and Combinations)

  1. #1
    Junior Member
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    Counting (Permutations and Combinations)

    I'm having a little trouble understanding some of this.

    This is an example from a practice test

    7 – 8. A poker hand consists of 5 cards dealt from a deck of 52 distinct cards.

    7. In how many ways a full house (3 of one type of card and 2 of another) be dealt?
    a. 168
    b. 432
    c. 3744
    d. 29,952
    e. none of these

    8. How many different hands contain exactly 3 aces?
    a. 1128
    b. 4512
    c. 1682
    d. 18,048
    e. none of these


    I have .. no idea how to approach this.

    There are 13 types of cards. So your chances of drawing a full house almost seems like it would be
    4*3*2 * 4*3

    But that doesn't make much sense either... gahh!
    Last edited by JTG2003; August 25th 2009 at 05:06 PM.
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  2. #2
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    Lexington, MA (USA)
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    [size=3]Hello, JTG2003!


    A poker hand consists of 5 cards dealt from a deck of 52 distinct cards.

    7. In how many ways a full house (3 of one type of card and 2 of another) be dealt?

    . . (a)\;168 \qquad(b)\;432 \qquad(c)\;3744\qquad(d)\;29,952\qquad (e)\;\text{none of these}
    We want a Triple and a Pair.

    There are 13 choices for the value of the Triple.
    There are: . {4\choose3} = 4 ways to get the Triple.

    There are 12 choices for the value of the Pair.
    There are: . {4\choose2} = 6 ways to get the Pair.

    Therefore, there are: . 13\cdot4\cdot12\cdot6 \:=\:{\color{blue}3744} ways to get a Full House.




    8. How many different hands contain exactly 3 aces?

    . . (a)\;1128  \qquad (b)\;4512  \qquad (c)\;1682 \qquad (d)\;18,\!048  \qquad (e)\;\text{none of these}
    The deck contains 4 Aces and 48 Others.
    . . We want 3 Aces and 2 Others.

    There are: . {4\choose3} = 4 ways to get 3 Aces.

    There are: . {48\choose2} = 1128 ways to get 2 Others.

    Therefore, there are: . 4\cdot1128 \:=\:{\color{blue}4512} ways to get exactly 3 Aces.

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