Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 9)
9.
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?
32
48
64
96
None of these
Answer: Option
Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
= 3 x 6 x 5 + 3 x 2 x 6 + 1
2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

Discussion:
81 comments Page 6 of 9.

Sanjay said:   1 decade ago
There is totally 9 balls and we can draw any 3 balls = 9c3 ways. A.q. at least one black ball always included and there is 3 black ball if exclude all the 3 black ball there is remaining 6 the no.of selection is 6c3.

The total no.of selection = 9c3-6c3.

= 64.

Chandu said:   1 decade ago
There is 9 balls is there of different colours from this 9 balls we select 3 so that we select: 9C3 ways and we donot considered 3 black balls,

So that we have 6 balls in this 6 balls we considered 3 balls so that we select : 6C3 ways and finally by subtracting from total selection - without containing black balls we get resulting answer: 10C3 - 6C3 = 64.

Saireshwanth said:   1 decade ago
According to the question we can take (1black+1white+1red) , (1black+2white) , (1black+2red) , (2black+1white) , (3black) by this way we can take (3c1*2c1*4c1) + (3c1*2c2) + (3c1*4c2) + (3c2*2c1) + (3c2*4c1) by this way also the answer is 64.

Sam said:   1 decade ago
If balls are IDENTICAL and ORDER does NOT matter - 6 ways.

If balls are IDENTICAL and ORDER matters - 19 ways.

If balls are NOT IDENTICAL - 64 ways.

Ernest said:   1 decade ago
Mani is correct.

Since White balls and Red balls are *indistinguishable*, as the text does not say otherwise, choosing red ball 1 and red ball 2 is the same as choosing red ball 2 and red ball 3, and these are not new combinations.

Hence:
B,W,R
B,W,W
B,R,R
B,B,R
B,B,W
B,B,B

Are the only options.

Shiva said:   1 decade ago
It means all wt ball is identical&all red balls are identical&all black balls are identical
Now we can solve it by negetive method i.e

Total 3 balls are selected in 9c3 ways
at least 1 black ball = at most 0 black balls i.e 6c3 ways

So the ans is 9c3-6c3=64

Biswajit said:   1 decade ago
In the question at least 1 black ball should be in that box.
So 3 black ball should not be selected.
ans- 63

Tamsha said:   1 decade ago
No 3 black is correct because dey have said there should b atleast 1 black ball. So you can take more then 1 but not less then 1. So eventually 3 is also correct. :).

Pasindu said:   1 decade ago
According to the question can't we choose nine balls in these ways
please anyone explain this

(1-black,2-white),(2-black,1-white) ,(1- black,2-red), (2-black,1-red),(1-black,1-white,1red)and (3-black)



(3-black) is this correct or wrong if this wrong please tell me why is that soon

Kranthi said:   1 decade ago
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


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