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


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