Class 11

Math

Algebra

Permutations and Combinations

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) at most $3$ girls?

This means you have to choose $3$ girls from $4$ and $4$ boys from $9$ to form a committee of $7$. This can be done in

$4C_{3}×9C_{4}$

(ii) At least $3$ girls

This means girls can either be $3$ or $4$

So first choose $3$ girls and $4$ boys or $4$ girls and $3$ boys

Total ways = $4C_{3}×9C_{4}$ + $4C_{4}×9C_{3}$

(iii) At most $3$ girls

This means we have to choose $1$ or $2$ or $3$ girls

So this can be done in

$4C_{1}×9C_{6}$ + $4C_{2}×9C_{5}$ + $4C_{3}×9C_{4}$ ways.