A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Answer: Option C

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
∴ Required number of ways
$$\begin{array}{rl}& =\left({}^3{C}_1\times {}^6{C}_2\right)+\left({}^3{C}_2\times {}^6{C}_1\right)+\left({}^3{C}_3\right)\\ & =\left(3\times \frac{6\times 5}{2\times 1}\right)+\left(\frac{3\times 2}{2\times 1}\times 6\right)+1\\ & =\left(45+18+1\right)\\ & =64\end{array}$$