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Math Help - Committees with 12 men and 8 women

  1. #1
    Member oldguynewstudent's Avatar
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    Committees with 12 men and 8 women

    I think I'm beginning to understand this, so I would appreciate if someone could critique the following:

    4. A group consists of 12 men and 8 women. How many ways are there to ...

    4.a) form a committee of size 5?
    \left({20\atop 5}\right)

    4.b) form a committee of size 5 containing two men and three women?
    \left({12\atop 2}\right)*\left({8\atop 3}\right)

    4.c) form a committee of size 6 containing at least three women?

    First approach: \left({12\atop 3}\right)*\left({8\atop 3}\right)+\left({12\atop 2}\right)*\left({8\atop 4}\right)+\left({12\atop 1}\right)*\left({8\atop 5}\right)+\left({12\atop 0}\right)*\left({8\atop 6}\right)

    Second approach: 2^{20}-\left({12\atop 4}\right)*\left({8\atop 2}\right)-\left({12\atop 5}\right)*\left({8\atop 1}\right)-\left({12\atop 6}\right)*\left({8\atop 0}\right)
    Last edited by oldguynewstudent; May 16th 2010 at 05:30 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by oldguynewstudent View Post
    I think I'm beginning to understand this, so I would appreciate if someone could critique the following:

    4. A group consists of 12 men and 8 women. How many ways are there to ...

    4.a) form a committee of size 5?
    \left({20\atop 5}\right)

    4.b) form a committee of size 5 containing two men and three women?
    \left({12\atop 2}\right)*\left({8\atop 3}\right)

    4.c) form a committee of size 6 containing at least three women?

    First approach: \left({12\atop 3}\right)*\left({8\atop 3}\right)+\left({12\atop 2}\right)*\left({8\atop 4}\right)+\left({12\atop 1}\right)*\left({8\atop 5}\right)+\left({12\atop 0}\right)*\left({8\atop 6}\right)

    Second approach: 12^{2}-\left({12\atop 4}\right)*\left({8\atop 2}\right)-\left({12\atop 5}\right)*\left({8\atop 1}\right)-\left({12\atop 6}\right)*\left({8\atop 0}\right)
    Your 4(c) 2nd approach is....

    subtract the committees with 4, 5 and 6 men from the total number of committees.

    This should have the three subtracted from \binom{20}{6} rather than 12^2
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  3. #3
    Member oldguynewstudent's Avatar
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    Thanks, I see how I really messed up the second approach.

    This help has been invaluable to me. Thanks for everything from everybody who has contributed.
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