The math department of a college has 20 faculty members, of whom 5 are women and 15 are men... Continued question: "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?" Possible Ways: * * * M M M M F F F F F F M M F F F M M M M F Is it suppose to be 2/5?

If there are more women than men, then there are either 3 women, 1 man or there are 4 women, 0 men.

If there are 3 women, 1 man, then there are a total of $\dbinom{5}{3} \times \dbinom{15}{1} = 150$ ways of doing it.

If there are 4 women, 0 men, then there are a total of $\dbinom{5}{4} \times \dbinom{15}{0} = 5$ ways of doing it.

Therefore, there are a total of $\boxed{155}$ ways of doing it.