In a committee of 5, which must be selected from 4 males and 3 females. In how many ways can the members be chosen if it were to include 2 females?
A
144 ways
B
15 ways
C
185 ways
D
12 ways
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
For the committee to include 2 females, we must have 3 males, so that there should be 5 members.
That is, \(^4C_3 \times ^3C_2\)
= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)
= 4 \times 3 = 12 ways
Source: Myschool Past Questions Archive (Record #55389)