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Math Help - probability of girls sit together/ seperated

  1. #1
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    probability of girls sit together/ seperated

    Ten Students sit in a row. Find the probability of 6 boys and 4 girls can be arranged if (i) all girls are side by side?
    (ii) all girls are seperated by boys


    my working for part i is 4!x7!=120960

    for part ii, why cant i just take 10! -120960=3507840 ? the ans given is 604800
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  2. #2
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    Re: probability of girls sit together/ seperated

    Hello, delso!

    Ten Students sit in a row.
    Find the probability of 6 boys and 4 girls can be arranged if

    (i) all girls are side by side?

    My working: .  7 \times 4!\times 6! \,=\,120,\!960
    Correct!



    (ii) all girls are seperated by boys.

    Why cant i just take 10! - 120,960=3,507,840? .This is wrong!

    The ans given is: 604,800.

    You subtracted the number of ways that all four girls are adjacent.
    What about "3 adjacent girls" or "2 adjacent girls"?


    Here is how I handle "not adjacent" problems.

    Place the 6 boys in a row.
    Insert a space before, after and between them.
    . . . \_\,b\,\_\,b\,\_\,b\,\_\,b\,\_\,b\,\_\,b\,\_
    There are 6! permutations of the boys.

    Choose 3 of the 7 spaces and assign the girls.
    There are _7P_4 choices.

    Therefore, there are: . (6!)(_7P_4) \:=\:(720)(840) \:=\:604,\!800 ways.
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