In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

Answer: Option D

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
∴ Required number of ways
$$=\left({}^6{C}_1{\times }^4{C}_3\right)+\left({}^6{C}_2{\times }^4{C}_2\right)+$$      $$\left({}^6{C}_3{\times }^4{C}_1\right)+$$   $$\left({}^6{C}_4\right)$$
$$=\left({}^6{C}_1{\times }^4{C}_1\right)+\left({}^6{C}_2{\times }^4{C}_2\right)+$$      $$\left({}^6{C}_3{\times }^4{C}_1\right)+$$   $$\left({}^6{C}_2\right)$$
$$=\left(6\times 4\right)+\left(\frac{6\times 5}{2\times 1}\times \frac{4\times 3}{2\times 1}\right)+$$       $$\left(\frac{6\times 5\times 4}{3\times 2\times 1}\times 4\right)+$$    $$\left(\frac{6\times 5}{2\times 1}\right)$$
$$\begin{array}{rl}& =\left(24+90+80+15\right)\\ & =209\end{array}$$