Correct Answer: (4)

Solution: (4)

Ways of selections $\Rightarrow$ (1 boy and 3 girls) or ( 2 boys and 2 girls) or ( 3 boys and 1 girl) or 4 boys.

$\therefore$ Required number of ways $=\left({ }^{6} C_{1} \times{ }^{4} C_{3}\right)+\left({ }^{6} C_{2} \times{ }^{4} C_{2}\right)+\left({ }^{6} C_{3}\right.$ $\left.\times{ }^{4} C_{1}\right)+{ }^{6} C_{4}$ $=\left({ }^{6} C_{1} \times{ }^{4} C_{1}\right)+\left({ }^{6} C_{2} \times{ }^{4} C_{2}\right)+\left({ }^{6} C_{3}\right.$

$\left.\times{ }^{4} C_{1}\right)+{ }^{6} C_{2}$ $=(6 \times 4)+\left(\frac{6 \times 5}{2 \times 1} \times \frac{4 \times 3}{2 \times 1}\right)+$

$\left(\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4\right)+\left(\frac{6 \times 5}{2 \times 1}\right)$ $=24+90+80+15=209$