In a committee of 5 which must be selected from... - JAMB Mathematics 2019 Question
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
correct option: d
For the committee to include 2 females, we must have 3 males, so that there should be a total of 5 members.
thus, \(^4C_3 \times ^3C_2\)
= \(\frac{4!}{(4 - 3)! 3!} \times \frac{3!}{(3 - 2)! 2!}\)
= 4 × 3
= 12 ways
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