(i) A committee consisting of 3 girls and 4 boys can be formed in ${ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4$ ways $={ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_4=\frac{4}{1} \times \frac{9 \times 8 \times 7 \times 6}{1.2 .3 .4}$ ways $=504$ ways
(ii) A committee having at least 3 girls will consists of
(a) 3 girls 4 boys, (b) 4 girls 3 boys This can be done in
$\begin{aligned}
&{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_4 \times{ }^9 \mathrm{C}_3 \text { ways } \\
&=\frac{4}{1} \times \frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4}+1 \times \frac{9 \times 8 \times 7}{1 \times 2 \times 3} \text { ways } \\
&\quad=504+84 \text { ways }=588 \text { ways }
\end{aligned}$
(iii) A committee having atmost 3 girls will consists of
(i) No girl, 7 boys (ii) 1 girls, 6 boys (iii) 2 girls, 5 boys (iv) 3 girls, 4 boys.
This can be done in
${ }^9 \mathrm{C}_7+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_6+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_5+{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4$ ways
${ }^9 \mathrm{C}_2+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_3+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_4$ ways
$=\frac{9 \times 8}{1 \times 2}+\frac{4}{1} \times \frac{9 \times 8 \times 7}{1 \times 2 \times 3}+\frac{4 \times 3}{1 \times 2} \times \frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4}$ $+\frac{4}{1} \times \frac{9 \times 8 \times 7 \times 6}{1 \times 2 \times 3 \times 4}$
$=36+4 \times 84+6 \times 126+4 \times 126$ ways
$=36+336+126(6+4)$ ways
$=36+336+1260$ ways $=1632$