Aptitude - Permutation and Combination - Discussion

Mani acoording to you,

If you take (1black+2red) or (1black+2 white) or (1black+1 red+1 white) or (2 black+ 1red) or (2black+1 white)

Then the answer will be

(3C1*4c2)+(3c1*2c2)+3c1*4c1*2c1)+(3c2*4c1)+(3c2*2c1)=64.

@ Mani: you are right in what you say but what you have answered is only half the question. what you have answered is in how many different combinations can the question requirements be satisfied.
(1 black and 2 white) = From those 9 balls you are specifically picking one black and two white balls here. The probability to do that would be 3/9 for the black or 1/3 then 2/8 or 1/4 for the first white ball (since a black is already removed) and 1/7 for the 2nd white ball. You will have to multiply those three numbers 1/3 x 1/4 x 1/7 to get the probability. By saying that the probability is 1 you are saying that there is a 100% guarantee that from those 9 balls you will always get 1 black and 2 white, which is obviously not the case. So, if you want to solve it this way I have done the first of the six combinations for you, you will need to do the other 5. This is a much longer method of course so you are better off doing the other method. :)

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 agree @Sam.

If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.

I have a doubt in this.

Its not mentioned in the question that all the balls are non-identical. It is said '3 black balls', so all the 3 black balls are identical.

The required no. of ways = (1 black and 2 white) or (1 black and 2 red) or (1 black and 1 white and 1 red) or (2 black and 1 white) or (2 black and 1 red) or (3 black)

= 1+1+1+1+1+1

= 6ways.

Can anyone please confirm my answer?

Thank u :)

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)
_______
64
_______

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.

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)?

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

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