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.
Kabilan said:
2 months ago
Why all three drawn can not be black balls in these cases why is it not included?
Adith said:
3 months ago
Very simple, here we have to first find the 9c3 - 6c3.
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)
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)
Emmy said:
2 years ago
I don't understand this, please. Can anyone help me out?
Ajle said:
2 years ago
Please explain 6c1 in detail.
Voldy said:
2 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?
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)
Shubham said:
4 years ago
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.
i.e,, 9C3-6C3.
(1)
Stige said:
4 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.
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