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.

AkashSingh said:   7 years ago
1 Black + 1 White + 1 Red.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
(1)

Tammunur Joshna said:   2 years ago
3C1 * 2C1 * 4C1 + 3C2 * 4C1 + 3C2*2C1 + 4C2*3C1 + 2C2*3C1 + 3C3.
= 3*2*4 + 3*4 + 3*2 + 6*3 + 3 +1.
= 64.
(8)

Rohit said:   1 decade ago
Guys why 'C' is taken in this. I could not understand please somebody can explain me in simple way.

Prasoon garg said:   10 years ago
I agree with @Mani as it is not mentioned in the question that all the balls are non-identical.

Bbbb said:   1 decade ago
@ Saireshwanth and Saurabh.

Could you explain how you get 64? Because I got the answer as 73.
(1)

Aditya Paras said:   11 months ago
@All.

Please Mention distinct balls, usually the same colour is used for similar objects
(4)

Kabilan said:   1 year ago
Why all three drawn can not be black balls in these cases why is it not included?

Sanjay Verma said:   8 years ago
3C1*6C2 which means they have merged red and white balls?

Am I right here?

Xyz said:   9 years ago
I agree @Vimal.

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

Sundar said:   1 decade ago
We should not take 3 black balls. Then why final declaration 3 ball added?


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