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Math Help - combinations

  1. #1
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    combinations

    Suppose that a department contains 12 men and 19 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by asugirl87 View Post
    Suppose that a department contains 12 men and 19 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?
    this means a committee can have:

    (a) 6 women and 0 men, or
    (b) 5 women and 1 man, or
    (c) 4 women and 2 men

    now find each of those and sum them up. do you know how to find each? (hint: think "combinations")
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  3. #3
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    C(6,0)*C(5,1)*C(4,2) is that what i use?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by asugirl87 View Post
    C(6,0)*C(5,1)*C(4,2) is that what i use?
    no

    for (b) for instance. there are C(19,5) ways to choose 5 women from 19 and C(12,1) ways to choose 1 man from 12. thus, for (b) there are C(19,5)*C(12,1) ways to form a committee with 5 women and 1 man

    now do the rest similarly
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