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:
78 comments Page 1 of 8.
Vish said:
2 years ago
We can also do it this way,
Atleast one black ball = total balls - No black ball
= 9c3 - 6c3.
9c3= 3 balls selected from the total no of balls,
6c3 - 3 balls taken from the total no of balls without any black which is 6(2 white + 3 red).
By calculating this, we will get the answer.
Atleast one black ball = total balls - No black ball
= 9c3 - 6c3.
9c3= 3 balls selected from the total no of balls,
6c3 - 3 balls taken from the total no of balls without any black which is 6(2 white + 3 red).
By calculating this, we will get the answer.
(12)
Tammunur Joshna said:
1 year ago
3C1 * 2C1 * 4C1 + 3C2 * 4C1 + 3C2*2C1 + 4C2*3C1 + 2C2*3C1 + 3C3.
= 3*2*4 + 3*4 + 3*2 + 6*3 + 3 +1.
= 64.
= 3*2*4 + 3*4 + 3*2 + 6*3 + 3 +1.
= 64.
(6)
Aditya said:
3 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)
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)
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)
Nilesh Sonune said:
5 years ago
Can it be like, 9C3-(2C2*4C1+2C1*4C2)= 68?
(1)
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)
Supratik said:
7 years ago
Why not 3C1 x 8C2=84 ?
(1)
Shubham said:
4 years ago
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.
i.e,, 9C3-6C3.
(1)
Anil said:
8 years ago
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?
(1)
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