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?

Why not 3C1*8c2?

I agree with @Mani as it is not mentioned in the question that all the balls are non-identical.

We should not take 3 black balls. Then why final declaration 3 ball added?

May be we can take (1B+1W+1R) OR (1B+2W) OR (1B+2R) OR (2B+1W) OR (2B+1R) OR all (3B).

So required no = (3C1*2C1*4C1)+(3C1*2C2)+(3C1*4C2)+(3C2*2C1)+(3C2*4C1)+3C3.

= (3*2*4)+(3*1)+(3*6)+(3*2)+(3*4)+1.

= 24+3+18+6+12+1 = 64.

Can't it be like 1 black ball is chosen from 3 black balls and then remaining 2 are chosen from 2 white, 4 red and remaining 2 black balls i.e 3.

= 3C1*8C2.

How you taken 6 in combination?

Here you can draw one black ball from x number of black balls in only one way, similarly 2 red balls from x numbers of red balls also in one way for xC2.

So, (3 Black) + (2 Black AND (1 Red OR 1 White)) + (1 Black AND ((2 white OR 2 Red OR (1 White AND 1 Red))).

-> 1 + 1 * 2 + 1 * (1+1+1).

-> 1 + 2 + 3.

-> 6.

Which I believe should be answer, any corrections or clarification most appreciated.

@ Saireshwanth and Saurabh.

Could you explain how you get 64? Because I got the answer as 73.

I have a doubt. In last question you have given nC(n-r). Why This conditions is not suitable for this?

In this question also why can't we do this as:

= (3C1 x 6C(6-2)) + (3C(3-2) x 6C1) + (3C3).