Aptitude - Permutation and Combination - Discussion

Consider no black ball is dran
there are totally 9 balls in which 3 balls can dran in 9c3 ways
we need to pick the 3 ball which are not black
we can do it in 6c3 ways

No of ways we can select is 9c3-6c3=64

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. :)

I think Pavel Sain's answer is correct.

I agree @Vimal.

Balls are to be considered as identical and as one object.

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 :)

It's simple.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.

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)?