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:
82 comments Page 3 of 9.
Shree said:
8 years ago
Thank you all for explaining the solution clearly.
Ria kedia said:
8 years ago
@Krish. Same doubt.
Amrish said:
9 years ago
@Sanjay
Answer should be 5 as balls are of the same color and hence the solution should be;
For 1 black ball- we could have 1 ball from either white or red lot. Or 2 from white and then 2 from red. So in total that makes 3 ways.
For 2 black balls- we can take 1ball from either lot in each condition. 2 ways.
For three black ball- just 1 way.
So total that makes 3+2+1= 5 ways.
Answer should be 5 as balls are of the same color and hence the solution should be;
For 1 black ball- we could have 1 ball from either white or red lot. Or 2 from white and then 2 from red. So in total that makes 3 ways.
For 2 black balls- we can take 1ball from either lot in each condition. 2 ways.
For three black ball- just 1 way.
So total that makes 3+2+1= 5 ways.
Sanjay Verma said:
9 years ago
3C1*6C2 which means they have merged red and white balls?
Am I right here?
Am I right here?
Aki47 said:
9 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.
Sourav said:
9 years ago
Why not this?
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
Kalai said:
9 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.
Fuji said:
9 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.
Naveen said:
9 years ago
I think it should be clearly specified whether balls of particular colour are identical or not.
If identical, then answer would be 6.
If identical, then answer would be 6.
Purvesh said:
8 years ago
Why is the difference between this 3c1 * 8c2=84.
And 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.
And 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.
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