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 5 of 9.
Pradeep said:
1 decade ago
@Anupam
9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final is also by 8 balls, that makes it 1*8*8=64.
Got it!
9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final is also by 8 balls, that makes it 1*8*8=64.
Got it!
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.
Sandeep said:
1 decade ago
Hi, at least one black ball is to be included in the draw.
So 3bc1*(2w+2b+4r)c2.
Why above answer is not correct? can any one explain that.
So 3bc1*(2w+2b+4r)c2.
Why above answer is not correct? can any one explain that.
Ankit said:
6 years ago
If I do like this;
(2C1*3C1*4C1)+(2C1*3C2*4C0)+(2C0*3C2*4C1)+(2C0*3C3*4C0). = (24+6+12+1)=43 WAYS.
Is this way right? Please tell me.
(2C1*3C1*4C1)+(2C1*3C2*4C0)+(2C0*3C2*4C1)+(2C0*3C3*4C0). = (24+6+12+1)=43 WAYS.
Is this way right? Please tell me.
(1)
Naveen said:
8 years ago
I think it should be clearly specified whether balls of particular colour are identical or not.
If identical, then answer would be 6.
If identical, then answer would be 6.
Vikash said:
9 years ago
I agree @Sam.
If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.
If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.
Himanshu Kulkarni said:
9 years ago
One black ball from 3 is 3C1 and any 2 balls from remaining 8 balls
i.e. 8C2.
So, why 3C1 x 8C2 does not match the solution?
i.e. 8C2.
So, why 3C1 x 8C2 does not match the solution?
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?
Krish said:
1 decade ago
3c1*8c2 = 84?
Why not? Atleast one black ball is there right? And from remainder we take 2 balls. What is wrong in this?
Why not? Atleast one black ball is there right? And from remainder we take 2 balls. What is wrong in this?
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
So 3 black ball should not be selected.
ans- 63
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