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Math Help - Answer in the back of the book wrong?

  1. #1
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    Answer in the back of the book wrong?

    Can someone please verify my answer? Because I'm sure I'm doing this question right but the answer at the back differs.

    Q: A bridge hand consists of 13 cards. How many bridge hands include 5 of one suit, 6 cards of a second, and 2 cards of a third?

    Isn't the solution: 13C5 x 13C6 x 13c2 = 172262376?

    The answer at the back is 4134297024

    EDIT:

    Oh I just realized that I have to account for the fact that only 3 of the 4 suits are being utilized. And order is important so do I multiply the 172 262 376 by 4P3?
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  2. #2
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    Hello, MATHDUDE2!

    Can someone please verify my answer?
    Because I'm sure I'm doing this question right but the answer at the back differs.

    Q: A bridge hand consists of 13 cards.
    How many bridge hands include 5 of one suit, 6 cards of a second, and 2 cards of a third?

    Isn't the solution: _{13}C_5 \times _{13}C_6 \times _{13}C_2 \:=\: 172,262,376 ?

    The answer at the back is:  4,134,297,024


    EDIT:
    Oh, I just realized that I have to account for the fact that only 3 of the 4 suits are being utilized.
    And order is important, so do I multiply the 17, 262,376 by _4P_3? . . . . Yes!
    Here's the way I approached it . . .

    There are {\color{blue}4} choices for the first suit.
    . . There are: {\color{blue}_{13}C_5} ways to get 5 cards of that suit.

    There are {\color{blue}3} choices for the second suit.
    . . There are: {\color{blue}_{13}C_6} ways to get 6 cards of that suit.

    There are {\color{blue}2} choices for the third suit.
    . . There are: {\color{blue}_{13}C_2} ways to get 2 cards of that suit.


    Therefore, the number of ways is: . \left(4 \times _{13}C_5\right) \times \left(3 \times _{13}C_6\right) \times \left(2 \times _{13}C_2\right)

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