Question

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?

Answer 1

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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

$C_{5,3}*C_{15,1} = \frac{5!}{3!(2)!}*\frac{15!}{1!(14)!} = 150$

4 women

4 women from a set of 5

$C_{5,4} = \frac{5!}{4!1!} = 5$

Total

150 + 5 = 155

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