Jambmaths question:
In how many ways can a committee of 2 women and 3 men be chosen from 6men and 5 women?
Jamb Maths Solution:
$\begin{align} & ^{6}{{C}_{3}}{{\times }^{5}}{{C}_{2}} \\ & \frac{6!}{3!(6-3)!}\times \frac{5!}{2!(5-2)!}=20\times 10=200 \\\end{align}$
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