Solve the problem.
-A class has 10 boys and 12 girls. In how many ways can a committee of four be selected if the Committee can have at most two girls?
A) 4410
B) 5170
C) 4620
D) 5665

Question

Quizplus · Accepted Answer

The total ways to form a committee of four with at most two girls can be calculated by considering all possible combinations: no girls, one girl, and two girls. - No girls (all boys): $C(10,4) = \frac{10!}{4!(10-4)!} = 210$- One girl: $C(12,1) \times C(10,3) = 12 \times \frac{10!}{3!(10-3)!} = 12 \times 120 = 1440$- Two girls: $C(12,2) \times C(10,2) = \frac{12!}{2!(12-2)!} \times \frac{10!}{2!(10-2)!} = 66 \times 45 = 2970$Adding these together: $210 + 1440 + 2970 = 4620$.

Solve the Problem. -A Class Has 10 Boys and 12 Girls. in How