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?
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.