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In how many ways can a committee of five people be selected from seven women and nine men if at least one woman must be in the committee?

  • math -

    From 7 women and 9 men = "16 people choose 5" for a committee, there are "16 choose 5" ways, or
    (16,5)=C(16,5)=16!/(5!(16-5)!) ways

    Out of these, "9 choose 5" ways include no woman, and must be subtracted.

    So the number of ways of selecting a committee of 5 from 9 men and 7 women with at least one woman is
    "16 choose 5" - "9 choose 5".



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