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 4 of 9.
Murali said:
8 years ago
If these are the 3 takes then one ball should be black and the remaining balls can be taken as randomly, so
1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.
1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.
Suvam said:
7 years ago
@All.
My question is why red & white colour balls are taken together ( by saying non-black) why we don't take it separately?
I think answer should be;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
My question is why red & white colour balls are taken together ( by saying non-black) why we don't take it separately?
I think answer should be;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
Ankit said:
7 years ago
@All.
My solution was:
(considering balls are non-identical)
1. Case(1 Blackball), Possibilities: 1*6*5 = 30.
2. Case (2 Black balls), Possibilities: 1*1*6 = 6.
3. Case(3 Black balls), Possibilities: 1*1*1 = 1.
Total ways: 36
I am not getting the answer, please anyone help me to get the right answer.
My solution was:
(considering balls are non-identical)
1. Case(1 Blackball), Possibilities: 1*6*5 = 30.
2. Case (2 Black balls), Possibilities: 1*1*6 = 6.
3. Case(3 Black balls), Possibilities: 1*1*1 = 1.
Total ways: 36
I am not getting the answer, please anyone help me to get the right answer.
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.
Nikolay said:
7 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:
7 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?
Voldy said:
4 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?
Emmy said:
3 years ago
I don't understand this, please. Can anyone help me out?
Kabilan said:
2 years ago
Why all three drawn can not be black balls in these cases why is it not included?
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