A committee of 7 has to be formed from 9 B and 4 G.
(i) exactly 3 girls
\(=^9C_4\times^4C_3\)
\(\frac{9!}{4!\space5!} \times \frac{4!}{3!\space1!}=\frac{9\times8\times7\times6\times5!}{3\times2\times5!}\)
\(=72\times7 = 504\)
(ii) at most 3 girls
(a) No girl and 7 boys
(b) 1 girl and 6 boys
(c) 2 girls and 5 boys
(d) 3 girls and 4 boys
∴ The committee consists of at most 3 girls
\(=^4C_0\times ^9C_7+^4C_1\times^9C_6+^4C_2\times^9C_5+^4C_3\times^9C_4\)
= 36 + 336 + 1296 + 504
= 2172