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 4 of 9.
Chiru said:
9 years ago
It's simple.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.
DEBRUPA SEN said:
6 years ago
Here the question is black balls are selected from 3 black balls. In solution, it is seen 6 black balls. How it is possible?
Nikolay said:
6 years ago
The total number of combinations can be 3x3x3 = 27 combinations of colours are possible.
But 1 out of 3 always have to be black so we exclude WWW and RRR, so this gives us 25 combinations - this has to be an answer. So, here option E is correct.
But 1 out of 3 always have to be black so we exclude WWW and RRR, so this gives us 25 combinations - this has to be an answer. So, here option E is correct.
Monika said:
6 years ago
We can say;
(Atleast one = Total - Zero).
So it's, 9C3- 3C0 * 6C3 = 64.
(Atleast one = Total - Zero).
So it's, 9C3- 3C0 * 6C3 = 64.
Anand said:
6 years ago
How (3C3)=1. Can anyone explain this, please?
Stige said:
5 years ago
(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
How to write this and (3C1 x 6C2) + (3C2 x 6C1)?
Please, anyone, explain.
How to write this and (3C1 x 6C2) + (3C2 x 6C1)?
Please, anyone, explain.
Voldy said:
3 years ago
Why is the following method giving wrong answers:
3C1*9C1*9C1 = 243.
But 3C1 for 1 black ball and other 2 9C1s for 2 balls of any colour.
Can someone say, why this logic is wrong?
3C1*9C1*9C1 = 243.
But 3C1 for 1 black ball and other 2 9C1s for 2 balls of any colour.
Can someone say, why this logic is wrong?
Ajle said:
3 years ago
Please explain 6c1 in detail.
Emmy said:
3 years ago
I don't understand this, please. Can anyone help me out?
Adith said:
1 year ago
Very simple, here we have to first find the 9c3 - 6c3.
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