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Math Help - Boys and Girls

  1. #1
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    Boys and Girls

    3 girls and 4 boys were standing in a circle. What is the probability that two girls are standing together, but the third one is not next to them?
    Last edited by Aquafina; November 1st 2009 at 09:25 PM.
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  2. #2
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    Question doesn't make any sense. One what is not with them?
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  3. #3
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    Quote Originally Posted by ANDS! View Post
    Question doesn't make any sense. One what is not with them?
    I have edited it to be clearer. This is what I did, there are 7 arrangements that 2 girls can stand in:

    GG***** the other girl can stand in 4 different places here
    *GG**** other girl in 3 places
    **GG*** 3
    ***GG** 3
    ****GG* 3
    *****GG 4

    So in total: 2x(4+3+3) = 20

    Is this correct? Do you have any other methods that you could show?
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  4. #4
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    Quote Originally Posted by Aquafina View Post
    I have edited it to be clearer. This is what I did, there are 7 arrangements that 2 girls can stand in:

    GG***** the other girl can stand in 4 different places here
    *GG**** other girl in 3 places
    **GG*** 3
    ***GG** 3
    ****GG* 3
    *****GG 4

    So in total: 2x(4+3+3) = 20

    Is this correct? Do you have any other methods that you could show?
    Dont forget that they are in a circle, so in your first case the third girl cannot take the position of the last astrix, as she would then be next to the first girl
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  5. #5
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    Quote Originally Posted by Robb View Post
    Dont forget that they are in a circle, so in your first case the third girl cannot take the position of the last astrix, as she would then be next to the first girl
    Ah thank you, so it should be:

    GG***** 3
    *GG**** 3
    **GG*** 3
    ***GG** 3
    ****GG* 3
    *****GG 3

    so 18 in total?

    Do you have any other methods?
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  6. #6
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    probability = no. ways you can arrange two girls together without third being adjacent / total no ways to arrange 3 girls and 4 boys in circle.

    B- boy
    G-girl
    BGGB - C

    First part is synomymous to arranging B,B,G,C on a straight line = 4!/2! = 12

    Second part = 7!/3!4! = 35

    Prob = 12/35
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