Aptitude - Permutation and Combination - Discussion

Consider the case where only one black ball is chosen. It could be chosen first second or third so that's 3 possibilities. As for the other two balls, they could be red or white. It could be red then white, white then red, white then white, or red then red, regardless of when the black ball is chosen.

So that's 4 possibilities. So, with only one black ball, that would be 12 possibilities (brw, bwr, brr, bww, rbw, wbr, rbr, wbw, rwb, wrb, rrb, wwb) but the solutions says there are 45 possibilities for this condition. Am I missing something or what?

One black ball from 3 is 3C1 and any 2 balls from remaining 8 balls
i.e. 8C2.

So, why 3C1 x 8C2 does not match the solution?

I think Pavel Sain's answer is correct.

@Pavel Sain: Since the ball of the same colour are identical, you should do ((3C1/3!) * (2C1/2!) * (4C1/4!)) etc.

So either you use the formula or make the individual combinations, the answer will be 6.

Answer is 64. If the colors are different then balls are not identical.

Can anybody tell me why the answer is not 3C1 x 8C2 = 84?

Can anybody tell me why the answer is not 3C1 x 8C2 = 84?

If we consider that the balls are not identical, then one black ball for sure can be selected in 3C1 ways, after that, you can select any 2 out of the remaining 8 in 8C2, which will include black also. Can someone explain why this 3C1 x 8C2 also equals the total number of ways of selecting 3 balls out of 9 (i.e. 9C3)?

There are 9 balls in box, and as given one of them has to be black so we get,
(1 black and any 2) + (2 black and any 1) + ( 3 black)
(3C1 x 8C2) + (3C2 x 7C1) + (3C3) = 168 + 21 + 3 = 192.

2 White balls (W)
3 black balls (B)
4 red balls (R)

1W 1R 1B : (2c1 * 3c1 * 4c1)
1W 2B : (2c1 * 3c2)
1R 2B : (4c1 * 3c2)
2W 1B : (2c2 * 3c1)
2R 1B : (4c2 * 3c1)
3B + (3c3)
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64
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