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.

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 = ${}^{4}\mathrm{C}_{3}\times {}^{9}\mathrm{C}_{4}$
$=\frac{4!}{3!\times 1!}\times \frac{9!}{5!\times 4!}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{4\times 3!}{3!}\times \frac{9\times 8\times 7\times 6\times 5!}{5!\times 4\times 3\times 2\times 1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=4\times 126\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=504$
(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 ${}^{4}\mathrm{C}_{3}\times {}^{9}\mathrm{C}_{4}$ ways.
4 girls and 3 boys can be selected in ${}^{4}\mathrm{C}_{4}\times {}^{9}\mathrm{C}_{3}$ways.
Therefore, in this case, required number of ways = ${}^{4}\mathrm{C}_{3}\times {}^{9}\mathrm{C}_{4}+{}^{4}\mathrm{C}_{4}\times {}^{9}\mathrm{C}_{3}$
= 504 + 84 = 588
(iii) Since atmost 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 ${}^{4}\mathrm{C}_{3}\times {}^{9}\mathrm{C}_{4}$ ways.
2 girls and 5 boys can be selected in ${}^{4}\mathrm{C}_{2}\times {}^{9}\mathrm{C}_{5}$ways.
1 girl and 6 boys can be selected in ${}^{4}\mathrm{C}_{1}\times {}^{9}\mathrm{C}_{6}$ways.
No girl and 7 boys can be selected in ${}^{4}\mathrm{C}_{0}\times {}^{9}\mathrm{C}_{7}$ways.
Therefore, in this case, required number of ways
$={}^{4}\mathrm{C}_{3}\times {}^{9}\mathrm{C}_{4}+{}^{4}\mathrm{C}_{2}\times {}^{9}\mathrm{C}_{5}+{}^{4}\mathrm{C}_{1}\times {}^{9}\mathrm{C}_{6}+{}^{4}\mathrm{C}_{0}\times {}^{9}\mathrm{C}_{7}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=504+6\times 126+4\times 84+1\times 36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=504+756+336+36\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=1632$