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) atleast 3 girls ? (iii) atmost 3 girls ?
(i) exactly 3 girls
Total numbers of girls are 4
Out of which 3 are to be chosen
∴ Number of ways in which choice would be made = 4C3
Numbers of boys are 9 out of which 4 are to be chosen which is given by 9C4
Total ways of forming the committee with exactly three girls
= 4C3 × 9C4
= 
(ii) at least 3 girls
There are two possibilities of making committee choosing at least 3 girls
There are 3 girls and 4 boys or there are 4 girls and 3 boys
Choosing three girls we have done in (i)
Choosing four girls and 3 boys would be done in 4C4 ways
And choosing 3 boys would be done in 9C3
Total ways = 4C4 ×9C3
Total numbers of ways of making the committee are
504 + 84 = 588
(iii) at most 3 girls
In this case the numbers of possibilities are
0 girl and 7 boys
1 girl and 6 boys
2 girls and 5 boys
3 girls and 4 boys
Number of ways to choose 0 girl and 7 boys = 4C0 × 9C7
Number of ways of choosing 1 girl and 6 boys = 4C1 × 9C6
Number of ways of choosing 2 girls and 5 boys = 4C2 × 9C5
Number of choosing 3 girls and 4 boys has been done in (1)
= 504
Total number of ways in which committee can have at most 3 girls are = 36 + 336 + 756 + 504 = 1632