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Math Help - Combinatorics Problem; Due Friday...

  1. #1
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    Jan 2008
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    Combinatorics Problem; Due Friday...

    I just can't seem to get this problem, and time is running out.

    The problems on this page all deal with hands taken from a deck of 32 cards:
    the 7, 8, ... , Q, K, A of each of the four suits.

    How many five-card hands are possible which satisfy:

    the hand has exactly two hearts
    the hand has exactly two spades
    the hand has exactly two Kings

    For this problem I set up the three cases like this: (O = non-heart, non-spade suits)

    1KH - 7H - 1KS - 7S - 14O
    1KH - 7H - 7S - 6S - 1KO
    7H - 6H - 1KS - 7S - 1KO
    Then I multiplied and added them together and got the
    correct answer of 1274.

    However, I can't seem to figure out the second and third
    problems.

    How many 6-card hands are possible which satisfy:
    the hand has exactly two hearts
    the hand has exactly three spades
    the hand has exactly 2 Kings

    I tried setting up the cases the same as in problem #1, with
    the necessary modifications:

    1KH - 7H - 1KS - 7S - 6S - 14O
    1KH - 7H - 7S - 6S - 5S - 1KO
    7H - 6H - 1KS - 7S - 6S - 1KO

    But My answer of 7350 is wrong and I can't figure out what
    I'm doing wrong.

    Similarly, I can't figure out the 3rd problem:

    How many 6-card hands are possible which satisfy:
    the hand has exactly two hearts
    the hand has exactly three spades

    I would think it would simply be:

    8H - 7H - 8S - 7S - 6S - 16O multiplied together, but it isn't.

    Any Help would be great.
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, shaehl!

    Surely, you're familiar with Combinations and Factorials ?!


    The problems on this page all deal with hands taken from a deck of 32 cards:
    the 7, 8, ... , Q, K, A of each of the four suits.

    1) How many five-card hands are possible which satisfy:
    . . (a) the hand has exactly two Hearts
    . . (b) the hand has exactly two Spades
    . . (c) the hand has exactly two Kings
    (a) Two Hearts

    There are 8 Hearts and 24 Others.
    We want 2 Hearts and 3 Others.

    To get 2 Hearts, there are: . {8\choose2} = 28 ways.
    To get 3 Others, there are: . {24\choose3} = 2,024 ways.

    Therefore: . n(2 \heartsuit) \:=\:28 \times 2,024 \:=\:\boxed{56,672}


    (b) Two Spades
    The reasoning (and the answer) is identical to part (a).


    (c) Two Kings

    There are 4 Kings and 28 Others.
    We want 2 Kings and 3 Others.

    To get 2 Kings, there are: . {4\choose2} \:=\:6 ways.
    To get 3 Others, there are: . {28\choose3} \:=\:3,276 ways.

    Therefore: . n(\text{2 Kings}) \:=\:6 \times 3,276 \:=\:\boxed{19,656}




    2) How many 6-card hands are possible which satisfy:
    . . (a) the hand has exactly two Hearts
    . . (b) the hand has exactly three Spades
    . . (c) the hand has exactly 2 Kings
    (a) Two Hearts

    There are 8 Hearts and 24 Others.
    We want 2 Hearts and 4 Others.

    To get 2 Hearts, there are: . {8\choose2} \:=\:28 ways.
    To get 4 Others, there are: . {24\choose4} \:=\:10,626 ways.

    Therefore: . n(2 \heartsuit) \:=\:28 \times 10,626 \:=\:\boxed{297,528}


    (b) Three Spades

    There are 8 Spades and 24 Others.
    We want 3 Spades and 3 Others.

    To get 3 Spades, there are: . {8\choose3} \:=\:56 ways.
    To get 3 Others, there are: . {24\choose3} \:=\:2,024 ways.

    Therefore: . n(2 \spadesuit) \:=\:56 \times 2,024 \:=\:\boxed{113,344}


    (c) Two Kings

    There are 4 Kings and 28 Others.
    We want 2 Kings and 4 Others.

    To get 3 Kings, there are: . {4\choose2} \:=\:6 ways.
    To get 4 Others, there are: . {28\choose4} \:=\:20.475 ways.

    Therefore: . n(\text{2 Kings}) \:=\:6 \times 20,475 \:=\:\boxed{122,850}




    3) How many 6-card hands are possible which satisfy:
    . . (a) the hand has exactly two Hearts
    . . (b) the hand has exactly three Spades
    Is there a typo?
    This is identical to problem #2.

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  3. #3
    Newbie
    Joined
    Jan 2008
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    Thank you for the thorough response, however, the problems are meant to satisfy all three of the conditions at once. For number 3, we exclude the 2 kings requirement.
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