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?

A committee of 7 has to be formed from 9 boys and 4 girls.

  1. Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.

Thus, in this case, required number of ways =  C34 × C49

= 4!3! × 1! × 9!5! × 4!=4×3!3! × 9×8×7×6×5!5! × 4×3×2×1=4 × 126=504

(ii) Since at least 3 girls are to be there in every committee, the committee can consist of

(a) 3 girls and 4 boys or (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in C34 × C49 ways.

4 girls and 3 boys can be selected in C44 × C39ways.

Therefore, in this case, required number of ways = C34 × C49 + C44 × C39 
= 504 + 84 = 588

(iii) Since at-most 3 girls are to be there in every committee, the committee can consist of

(a) 3 girls and 4 boys (b) 2 girls and 5 boys

(c) 1 girl and 6 boys (d) No girl and 7 boys

3 girls and 4 boys can be selected in C34 × C49 ways.

2 girls and 5 boys can be selected in C24 × C59ways.

1 girl and 6 boys can be selected in C14 × C69ways.

No girl and 7 boys can be selected in C04 × C79ways.

Therefore, in this case, required number of ways

 = C34 × C49 + C24 × C59 + C14 × C69 + C04 × C79=504 + 6×126 + 4×84 + 1×36=504 + 756 + 336 + 36=1632

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