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 2 of 9.
Aditya said:
4 years ago
@All.
The Alternate way for the solution is:
All ways of picking 3 balls - All ways in which no black ball is picked = All ways in atleast 1 black ball is pick;
9c3 - 6c3 = 84 - 20 = 64.
The Alternate way for the solution is:
All ways of picking 3 balls - All ways in which no black ball is picked = All ways in atleast 1 black ball is pick;
9c3 - 6c3 = 84 - 20 = 64.
(1)
Shubham said:
5 years ago
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.
i.e,, 9C3-6C3.
(1)
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.
Tejas said:
5 years ago
Why the following method is wrong? Please tell me.
He have total 9 balls out of which 3 to be selected.
out of 3, one must be black.. out of 3 black we select 1. Now we have 8 balls remaining and a total of 2 places hence 8c2=28.
He have total 9 balls out of which 3 to be selected.
out of 3, one must be black.. out of 3 black we select 1. Now we have 8 balls remaining and a total of 2 places hence 8c2=28.
(1)
Nilesh Sonune said:
5 years ago
Can it be like, 9C3-(2C2*4C1+2C1*4C2)= 68?
(1)
Anand said:
6 years ago
How (3C3)=1. Can anyone explain this, please?
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.
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.
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)
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?
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