Question

In how many ways can a committee of nine be divided into two subcommittee, one having five members and the other having four?

Answer 1

$$5+4=9$$

$$\dbinom{9}{5}=\dfrac{9!}{5!(9-5)!}=126$$

$$\dbinom{9}{4}=\dfrac{9!}{4!(9-4)!}=126$$

A committee of nine can be divided into two subcommittee, one having five members and the other having four in 126 ways.