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:
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)!)](https://img.qammunity.org/2021/formulas/mathematics/college/qaowm9lzn4vyb0kbgc2ooqh7fbldb6dkwq.png)
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](https://img.qammunity.org/2021/formulas/mathematics/college/hon3pye1o6nom6vkmxhgg1rer8a1s3an8k.png)
4 women
4 women from a set of 5
![C_(5,4) = (5!)/(4!1!) = 5](https://img.qammunity.org/2021/formulas/mathematics/college/hakoy38mlg6yt22nm62kdr9emgiwbd5njf.png)
Total
150 + 5 = 155
155 ways are there to select the committee that has more women than men.