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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 & Solution
Correct 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
$$\eqalign{ & = \left( {{}^3{C_1} \times {}^6{C_2}} \right) + \left( {{}^3{C_2} \times {}^6{C_1}} \right) + \left( {{}^3{C_3}} \right) \cr & = \left( {3 \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {\frac{{3 \times 2}}{{2 \times 1}} \times 6} \right) + 1 \cr & = \left( {45 + 18 + 1} \right) \cr & = 64 \cr} $$
∴ 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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