Question

Suppose that a department contains 9 men and 15 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?

Answer 1

Answer:  The required number of different possible committees is 81172.

Step-by-step explanation:    Given that a department contains 9 men and 15 women.

We are to find the number of different committees of 6 members that are possible if the committee must have strictly more women than men.

Since we need committees of 6 members, so the possible combinations are

(4 women, 2 men), (5 women, 1 men) and (6 women).

Therefore, the number of different committees of 6 members is given by

$n\\\\\\=^{15}C_4\times ^9C_2+^{15}C_5\times ^9C_1+^{15}C_6\\\\\\=\dfrac{15!}{4!(15-4)!}\times \dfrac{9!}{2!(9-2)!}+\dfrac{15!}{5!(15-5)!}\times \dfrac{9!}{1!(9-1)!}+\dfrac{15!}{6!(15-6)!}\\\\\\\\=\dfrac{15\times14\times13\times12\times11!}{4\times3\times2\times1\times11!}\times\dfrac{9\times8\times7!}{2\times1\times7!}+\dfrac{15\times14\times13\times12\times11\times10!}{5\times4\times3\times2\times1\times10!}\times\dfrac{9\times8!}{1\times8!}+\dfrac{15\times14\times13\times12\times11\times10\times9!}{6\times5\times4\times3\times2\times1\times9!}\\\\\\=1365\times36+3003\times9+5005\\\\=49140+27027+5005\\\\=81172.$

Thus, the required number of different possible committees of 6 members is 81172.