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Suppose that a department contains 8 men and 20 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

  • statistics -

    If more women than men, possibilities are
    6W, 0M
    5W, 1M
    4W, 2M
    3W, 3M --- no longer possible

    6W,0M ---> C(20,6) x C(8,0) = 38760
    5W, 1M ---> C(20,5) x (8,1) = 15504(8) = 124032
    4W, 2M ---> C(20,4) x C(8,2) = 4845(28) = 135660

    add them up

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