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Math Help - Extremely Simple Circle Probability Problem! :)

  1. #1
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    Extremely Simple Circle Problem! :)

    How many ways can 8 boys be arranged in a circle if 3 boys cannot sit next to each other?

    Well, I don't actually have the correct answer, but this is how I solved it:

    Total Ways the Boys Can Be Arranged - 3 Boys Sitting Together = Ways The 3 Boys Cannot Sit Together

    So, I did:

    (8-1)! - (5! x 3!) = 4320 ways

    I think this is wrong though. If someone would be able to solve it and explain the reasoning behind it, that would be awesome!
    Last edited by saintv; December 15th 2009 at 07:18 PM.
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  2. #2
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    Hello, saintv!


    How many ways can 8 boys be arranged in a circle
    if 3 boys cannot sit next to each other?

    This is how I solved it:

    (Total ways boys can be arranged) - (3 boys together) .= .(Ways 3 boys are not together)

    So, I did: . (8-1)! - (5!\cdot 3!) \:=\: 4320 ways

    I think this is wrong though.
    It depends on the interpretation of "the 3 boys are not together".

    Suppose the eight boys are: {A, B, C, D, E, X, Y, Z}
    . . and X, Y, Z must not sit in adjacent chairs.


    If "the 3 boys are together" means XYZ (in some order),
    . . your solution is correct.
    Note that this allows, for example: . X A Y Z B C D E
    . . where two of them can be together.


    If no two of those boys can be adjacent, a different approach is used.

    Place the other 5 in a circle; there are: . (5-1)!=24 ways.
    . . There are 5 spaces between these 5 boys.
    Select 3 of these spaces to place X, Y, Z.
    . . There are: . _5P_3 = 60 ways.

    Therefore, there are: . 24 \times 60 \,=\,1440 ways.

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