In how many ways can a committee of 2 women and... - JAMB Mathematics 2010 Question

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