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 4 of 9.
Aditya said:
4 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)
Voldy said:
3 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?
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.
Fuji said:
8 years ago
I selected 1 ball from 3 black ball. That reduces it to 2. So I get 2+2+4 balls of white, black and red respectively. So I get 8c2 ways. Only 59.
Am I correct? Please Explain.
Am I correct? Please Explain.
Tamsha said:
1 decade ago
No 3 black is correct because dey have said there should b atleast 1 black ball. So you can take more then 1 but not less then 1. So eventually 3 is also correct. :).
Aki47 said:
8 years ago
Why not this?
1 black ball = 3 ways.
Remaining are 4+2+2 = 8 balls.
Remaining 2 can be chooses in 8 x 7 = 56 ways.
No. of combinations thus = 3 x 56 = 168 ways.
1 black ball = 3 ways.
Remaining are 4+2+2 = 8 balls.
Remaining 2 can be chooses in 8 x 7 = 56 ways.
No. of combinations thus = 3 x 56 = 168 ways.
Tanya said:
1 decade ago
Can't it be like 1 black ball is chosen from 3 black balls and then remaining 2 are chosen from 2 white, 4 red and remaining 2 black balls i.e 3.
= 3C1*8C2.
= 3C1*8C2.
(1)
Kalai said:
8 years ago
Hi @Fuji.
You take 3 balls at a time and that's all. You don't keep it aside. Every time you take is a new combination. I hope you are understanding it.
You take 3 balls at a time and that's all. You don't keep it aside. Every time you take is a new combination. I hope you are understanding it.
Chiru said:
9 years ago
It's simple.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.
Sam said:
1 decade ago
If balls are IDENTICAL and ORDER does NOT matter - 6 ways.
If balls are IDENTICAL and ORDER matters - 19 ways.
If balls are NOT IDENTICAL - 64 ways.
If balls are IDENTICAL and ORDER matters - 19 ways.
If balls are NOT IDENTICAL - 64 ways.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers