Question

a) How many ways can a teacher form a committee of 3 members from a class of 4 boys and 3 girls?

b)How many committees in (a) has exactly one girl?

c)How many committees in (a) has at least one girl?

Answer 1

a) N=4+3=7

n=3

$Number \;of \;ways = \frac{N!}{n!(N-n)!} \\ = \frac{7!}{3!(7-3)!} \\ = \frac{5 \times 6 \times 7}{2 \times 3} \\ = 35$

b) When the number of girls in a committee is exactly 1, means the number of boys in the committee should be 2.

Number of combinations having 1 girl and 2 boys in committe $= С(3,1) \times C(4,2)$

$= 3 \times 6 = 18$

c) Number of combinations containing only boys $= C(4,3) = \frac{4!}{3!(4-3)!} = \frac{4}{1}=4$

Number of combinations containing at least one girl = 35-4 = 31