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:
81 comments Page 8 of 9.
Anand said:
6 years ago
How (3C3)=1. Can anyone explain this, please?
Nilesh Sonune said:
5 years ago
Can it be like, 9C3-(2C2*4C1+2C1*4C2)= 68?
(1)
Revathy said:
9 years ago
I think Pavel Sain's answer is correct.
Jumenal said:
1 decade ago
How you taken 6 in combination?
(1)
Ajle said:
3 years ago
Please explain 6c1 in detail.
Prince said:
8 years ago
6C2, 6C1..why we take 6 here?
Supratik said:
8 years ago
Why not 3C1 x 8C2=84 ?
(1)
Ashu Pathak said:
7 years ago
Why not 3c1 * 8c2=84?
(1)
Pavi said:
3 months ago
i want simple explain
Ria kedia said:
8 years ago
@Krish. Same doubt.
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