A committee of 7 has to be formed from 9 boys and 4 girls.
(i) 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 = \(^4C_3\times\space^9C_4=\frac{4!}{3!1!}\times\frac{9!}{4!5!}\)
\(=4\times\frac{9\times8\times7\times5!}{4\times3\times2\times1\times5!}\)
\(=504\)
(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 \(^4C_3\times\space^9C_4\) ways.
4 girls and 3 boys can be selected in \(^4C_4\times \space^9C_3\) ways.
Therefore, in this case, required number of ways = \(^4C_3\times\space^9C_4+^4C_4\times\space^9C_3\)
\(=504+84=588\)
(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 \(^4C_3\times\space^9C_4\) ways.
2 girls and 5 boys can be selected in \(^4C_2\times\space^9C_5\) ways.
1 girl and 6 boys can be selected in \(^4C_1\times\space^9C_6\) ways.
No girl and 7 boys can be selected in \(^4C_0\times\space^9C_7\) ways.
Therefore, in this case, required number of ways
\(=\)\(^4C_3\times\space^9C_4+^4C_2\times^9C_5+^4C_1\times\space^9C_6+^4C_0\times\space^9C_7\)
\(=\)\(\frac{4!}{3!1!}\times\frac{9!}{4!5!}\)\(+\frac{4!}{2!2!}\times\frac{9!}{5!4!}\)\(+\frac{4!}{1!3!}\times\frac{9!}{6!3!}+\frac{4!}{0!4!}\times\frac{9!}{7!2!}\)
\(=504+756+336+36\)
\(=1632\)