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The math department of a college has 20 faculty members, of whom 5 are women and 15 are men. A curriculum committee of 4 faculty members is to be selected. How many ways are there to select the committee that has more women than men?

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Answer:

155 ways are there to select the committee that has more women than men.

Explanation:

The order is not important.

For examle, a committee of John, Elisa, Josh and Rose is the same committee as Elisa, John, Josh and Rose. So we use the combinations formula to solve this problem.

Combinations formula:


C_(n,x) is the number of different combinations of x objects from a set of n elements, given by the following formula.


C_(n,x) = (n!)/(x!(n-x)!)

How many ways are there to select the committee that has more women than men?

It can either be 3 women and a men, or 4 women.

3 women and a men

1 men from a set of 15

3 women from a set of 5


C_(5,3)*C_(15,1) = (5!)/(3!(2)!)*(15!)/(1!(14)!) = 150

4 women

4 women from a set of 5


C_(5,4) = (5!)/(4!1!) = 5

Total

150 + 5 = 155

155 ways are there to select the committee that has more women than men.

User Conley Owens
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