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?
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) | |||||||||||||
|
||||||||||||||
| = (45 + 18 + 1) | ||||||||||||||
| = 64. |
Discussion:
82 comments Page 9 of 9.
Vibhuti said:
8 months ago
Why not permuting balls? Please explain to me.
Mariyada Ramesh said:
2 weeks ago
(3C1 x 6C2) + (3C2 x 6C1) + (3C3).
In this problem, how they got 6?
I'm not understanding, please explain to me.
In this problem, how they got 6?
I'm not understanding, please explain to me.
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