Number of girls $=4$ and Number of boys $=7$
We have to select a team of 5 members
(i) When the team has no girls.
$\therefore$ Required number of ways
$$
={ }^7 C_5=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6}{2}=21
$$
(ii) When the team has atleast one boy and one girl
$\therefore$ Required number of ways
$$
\begin{aligned}
&={ }^7 C_1 \times{ }^4 C_4+{ }^7 C_2 \times{ }^4 C_3+{ }^7 C_3 \times{ }^4 C_2+{ }^7 C_4 \times{ }^4 C_1 \\
&=7 \times 1+21 \times 4+35 \times 6+35 \times 4 \\
&=7+84+210+140=441
\end{aligned}
$$
(iii) When the team has atleast three girls
$\therefore$ Required number of ways
$$
\begin{aligned}
&={ }^4 C_3 \times{ }^7 C_2+{ }^4 C_4 \times{ }^7 C_1 \\
&=4 \times 21+7=84+7=91
\end{aligned}
$$