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Math Help - number of ways

  1. #1
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    number of ways

    The question is:

    In how many ways can 8 different CD's be distributed to 3 people so that each person receives at least 2 CD's?

    My solution:

    (8C2 * 6C3 * 3C3) * 3! = 3360

    (number of ways that two people get 3 CD's and one person gets 2 CD's)

    (8C2 * 6C2 * 4C4) * 3! = 2520

    (number of ways that two people get 2 CD's and one people gets four CD's)

    3360+2520= 5880


    However, the correct answer is 6720. What have I done wrong? Any help is very much appreciated. Thanks
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, justinc89!

    Your factorial is incorrect.
    . . And I don't agree with their answer . . .


    In how many ways can 8 different CD's be distributed to 3 people,
    so that each person receives at least 2 CD's?

    My solution:

    (8C2 * 6C3 * 3C3) * 3! = 3360
    . .
    (number of ways that two people get 3 CD's and one person gets 2 CD's)

    (8C2 * 6C2 * 4C4) * 3! = 2520
    . .
    (number of ways that two people get 2 CD's and one people gets four CD's)

    3360 + 2520 = 5880

    However, the correct answer is 6720. ?
    You should use 3 instead of 3!.

    To divide them into (2,3,3), there are: . {8\choose2}{6\choose3}{3\choose3} \:=\:560 ways.
    But there are not 3! permutations; there are only 3.
    . . [The decision is: Who gets the "2"? . . . There are 3 choices.)
    So there are: . 3 \times 560 \:=\:1680 ways for (2,3,3).

    To divide them into (2,2,4), there are: . {8\choose2}{6\choose2}{4\choose4} \:=\:420 ways.
    The decision is: Who gets the "4"? . . . There are 3 choices.
    So there are: . 3 \times 420 \:=\:1260 ways for (2,2,4).


    Therefore, there are: . 1680 + 1260 \:=\:\boxed{2940} ways.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Their answer is way too big.

    With no restrictions, there are only: . 3^8 \:=\:6561 ways.

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  3. #3
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    Joined
    Apr 2008
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    Thanks a lot, I figured the answer key was wrong.
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