Question on: JAMB Mathematics - 2010
In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?
A
100
B
200
C
30
D
50
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Correct Option: B
A committee of 2 women and 3 men can be chosen from 6 men and 5 women, in (^{5}C_{2}) x (^{6}C_{3}) ways
= (\frac{5!}{(5 - 2)!2!} \times {\frac{6!}{(6 - 3)!3!}})
= (\frac{5!}{3!2!} \times {\frac{6!}{3 \times 3!}})
= (\frac{5 \times 4 \times 3!}{3! \times 2!} \times {\frac{6 \times 5 \times 4 \times 3!}{3! \times 3!}})
= (\frac{5 \times 4}{1 \times 2} \times {\frac{6 \times 5 \times 4}{1 \times 2 \times 3}})
= 10 x (\frac{6 \times 20}{6})
= 200
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