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JAMB Mathematics 2010 Paper

Question : 45

Total: 49

Solution:

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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