**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) at most 3 girls?

**Solution:**

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

I. Exactly 3 girls: If there are exactly 3 girls in the committee, then there must be 4 boys, and the ways in which they can be chosen is

$={ }^{4} \mathrm{C}_{3}$ $x$ ${ }^{9} \mathrm{C}_{4}$

II. At least 3 girls: Here the possibilities are

(i) 3 girls and 4 boys and

(ii) 4 girls and 3 boys.

The number of ways they can be selected

$={ }^{4} \mathrm{C}_{3}$ $\times$ ${ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{4}$ $\times$ ${ }^{9} \mathrm{C}_{3}$

$=588$

III. At most 3 girls:

(i) 7 boys but no girls

(ii) 6 boys and 1 girl

(iii) 5 boys and 2 girls \&

(iv) 4 boys and 3 girls.

And the number of their selection is

$={ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}$

$=1584$ ways.

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