math
posted by Fakaapo .
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?

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!(165)!) 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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