Answer to Question #330172 in Statistics and Probability for Robera

Question

In how many ways can a committee of three be chosen from 4 married couples if

a) All are equally eligible?

b) One particular man must be on the committee?

Answer 1

a) $N=C_8^3=\frac{8!}{3!5!}=56.$

b) $N=C_7^2=\frac{7!}{2!5!}=21.$