Question

A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:

(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?

Answer 1

(i) There are 9 boys and 4 girls. We have to select exactly 3 girls out of 4 girls and 4 boys out of 9 boys.

$\therefore$ Number of ways of selection = ${}^9{C}_4{\times }^4{C}_3=\frac{9!}{4!5!}\times \frac{4!}{3!1!}$

= $\frac{9\times 8\times 7\times 6\times 5!}{4!5!}\times \frac{4!}{3\times 2\times 1}=504$

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.

$\therefore$ Number of ways of selection = ${}^4{C}_3{\times }^9{C}_4{+}^4{C}_4{\times }^9{C}_3$

= $\frac{4!}{3!1!}\times \frac{9!}{4!5!}+\frac{4!}{4!0!}\times \frac{9!}{3!6!}$

= $\frac{4!}{3\times 2\times 1}\times \frac{9\times 8\times 7\times 6\times 5!}{4!5!}+1\times \frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}$

= $504+84=588.$

(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

$\therefore$ Number of ways of selection

= ${}^4{C}_0{\times }^9{C}_7{+}^4{C}_1{\times }^9{C}_6{+}^4{C}_2{\times }^9{C}_5{+}^4{C}_3{\times }^9{C}_4$

= $1\times \frac{9!}{7!2!}+\frac{4!}{1!3!}\times \frac{9!}{6!3!}+\frac{4!}{2!2!}\times \frac{9!}{5!4!}+\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

= $1\times \frac{9\times 8\times 7!}{7!\times 2\times 1}+\frac{4\times 3!}{1\times 3!}\times \frac{9\times 8\times 7\times 6!}{6!\times 3\times 2\times 1}+\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{9\times 8\times 7\times 6\times 5!}{5!\times 4\times 3\times 2\times 1}+\frac{4\times 3!}{3!\times 1}\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}$

= $36+336+756+504=1632.$