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?

A. 32

B. 48

C. 64

D. 96

E. None of these

Answer: Option C

Solution(By Examveda Team)

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
∴ Required number of ways
$$\eqalign{ & = \left( {{}^3{C_1} \times {}^6{C_2}} \right) + \left( {{}^3{C_2} \times {}^6{C_1}} \right) + \left( {{}^3{C_3}} \right) \cr & = \left( {3 \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {\frac{{3 \times 2}}{{2 \times 1}} \times 6} \right) + 1 \cr & = \left( {45 + 18 + 1} \right) \cr & = 64 \cr} $$