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 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways  = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{3}C_{3})  


= (45 + 18 + 1)  
= 64. 
Discussion:
78 comments Page 2 of 8.
Tejas said:
4 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:
5 years ago
How (3C3)=1. Can anyone explain this, please?
Monika said:
5 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:
5 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:
5 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:
5 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?
Ankit said:
5 years ago
@All.
My solution was:
(considering balls are nonidentical)
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 nonidentical)
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.
Suvam said:
6 years ago
@All.
My question is why red & white colour balls are taken together ( by saying nonblack) 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 nonblack) why we don't take it separately?
I think answer should be;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
AkashSingh said:
6 years ago
1 Black + 1 White + 1 Red.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
(1)
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