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A box contains 2 white, 3 black and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least 1 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
$$\eqalign{ & \left( {{}^3{C_1} \times {}^6{C_2}} \right) + \left( {{}^3{C_2} \times {}^6{C_1}} \right) + \left( {{}^3{C_3}} \right) \cr & = \left\{ {3 \times \frac{{6 \times 5}}{{2 \times 1}}} \right\} + \left( {\frac{{3 \times 2}}{{2 \times 1}} \times 6} \right) + 1 \cr & = \left( {45 + 18 + 1} \right) \cr & = 64 \cr} $$

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