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 7 of 9.
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.
Apluv said:
9 years ago
Answer is 64. If the colors are different then balls are not identical.
Shubham said:
5 years ago
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.
i.e,, 9C3-6C3.
(1)
Gauss said:
9 years ago
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?
Anil said:
9 years ago
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?
(1)
Emmy said:
3 years ago
I don't understand this, please. Can anyone help me out?
Adith said:
1 year ago
Very simple, here we have to first find the 9c3 - 6c3.
Shree said:
7 years ago
Thank you all for explaining the solution clearly.
Sourav said:
8 years ago
Why not this?
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
Vibhuti said:
1 month ago
Why not permuting balls? Please explain to me.
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