Number of girls $=4$ and Number of boys $=7$

We have to select a team of 5 members provided that

(i) team having no girls.

$$\therefore \quad \text { Required selection }={ }^7 C_5=\frac{7!}{5!2!}=\frac{7 \times 6}{2}=21$$

(ii) atleast one boy and one girl

$$\begin{aligned} \therefore \text { Required selection } & ={ }^7 C_1 \times{ }^4 C_4+{ }^7 C_2 \times{ }^4 C_3+{ }^7 C_3 \times{ }^4 C_2+{ }^7 C_4 \times{ }^4 C_1 \\ & =7 \times 1+21 \times 4+35 \times 6+35 \times 4 \\ & =7+84+210+140=441 \end{aligned}$$

(iii) when atleast three girls are included $={ }^4 C_3 \times{ }^7 C_2+{ }^4 C_4 \times{ }^7 C_1$

$$=4 \times 21+7=84+7=91$$

$\because\quad$ Total number of men = 10

and total number of women =7

We have to form a committee containing atleast 3 men and 2 women.

Number of ways $={ }^{10} \mathrm{C}_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2$

If two particular women to be always there .

$\therefore \quad$ Number of ways $={ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1$

Total number of committee when two particular women are never together

$$\begin{aligned} & =\text { Total }- \text { Together } \\ & =\left({ }^{10} C_3 \times{ }^7 C_3+{ }^{10} C_4 \times{ }^7 C_2\right)-\left({ }^{10} C_4 \times{ }^5 C_0+{ }^{10} C_3 \times{ }^5 C_1\right) \\ & =(120 \times 35+210 \times 21)-(210+120 \times 5) \\ & =4200+4410-(210+600) \\ & =8610-810=7800 \end{aligned}$$