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?

6

, if the balls of the same colour are alike

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8

, if the balls of the same colour are alike

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64

, if the balls of the same colour are different.

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80

, if the balls of the same colour are different.

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Solution

The correct options are
B $64$ , if the balls of the same colour are different.
D $6$ , if the balls of the same colour are alike
1. If the balls of the same colour are identical then the following combinations are possible:
$(B W R), (B W W), (B R R), (B B W), (B B R), (B B B)$
So, totally $6$ combinations are possible in the above case.
2. If balls of the same colour are different,
Total combination = $^{9} C_{3} = 84$ ways
We need only the cases where atleast 1 black ball is selected.
Number of such cases $= 84 -$ (combinations where there is no black ball)= $84 - (^{6} C_{3}) = 84 - 20 = 64$ ways
Hence, options 'A' and 'C' are correct.

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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?

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?