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 5 of 9.
Kabilan said:
1 year ago
Why all three drawn can not be black balls in these cases why is it not included?
Pavi said:
3 months ago
i want simple explain
Vibhuti said:
1 month ago
Why not permuting balls? Please explain to me.
Garry said:
1 decade ago
I believe Mani is correct. Consider this: From 2 red balls, you must choose 1. How many ways are there of doing this ?
From two balls, you can choose 1, 2C1 or 2. But there is only one possible outcome: {1 red ball}. Since the two red balls are identical, there is actually only 1 way to choose, not 2.
So there are actually 6 ways of choosing as Mani said, not 64.
From two balls, you can choose 1, 2C1 or 2. But there is only one possible outcome: {1 red ball}. Since the two red balls are identical, there is actually only 1 way to choose, not 2.
So there are actually 6 ways of choosing as Mani said, not 64.
Prateek said:
1 decade ago
Why 3c1*8c2 doesn't give the right answer...in 3c1 you select any one of the 3 black balls...then in in 8c2 you select any 2 balls out of remaining 8 balls which may or may not include other black balls.....in this way you will always have at least one ball.
Also 3c1 * 8c2 = 9c3... no idea why?
Also 3c1 * 8c2 = 9c3... no idea why?
Pradeep said:
1 decade ago
@Anupam
9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final is also by 8 balls, that makes it 1*8*8=64.
Got it!
9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final is also by 8 balls, that makes it 1*8*8=64.
Got it!
Anupam Patra said:
1 decade ago
Why can't I do like this.
One black ball is compulsory, So 9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final by 7 balls, that makes it 1*8*7=56.
One black ball is compulsory, So 9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final by 7 balls, that makes it 1*8*7=56.
Rohit said:
1 decade ago
Guys why 'C' is taken in this. I could not understand please somebody can explain me in simple way.
Vimal said:
1 decade ago
According to your point, the combination are 1 black and 2 white or 1 black and 2 red or 3 black. The other combination such as (1 black and 1w and 1r),(2b and 1w) or (2b and 1r) are missed here. So the answer is wrong for your assumption.
I hope you understood.
I hope you understood.
Sandeep said:
1 decade ago
Hi, at least one black ball is to be included in the draw.
So 3bc1*(2w+2b+4r)c2.
Why above answer is not correct? can any one explain that.
So 3bc1*(2w+2b+4r)c2.
Why above answer is not correct? can any one explain that.
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