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 6 of 9.
Sanjay Verma said:
8 years ago
3C1*6C2 which means they have merged red and white balls?
Am I right here?
Am I right here?
Amrish said:
8 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.
Ria kedia said:
8 years ago
@Krish. Same doubt.
Supratik said:
8 years ago
Why not 3C1 x 8C2=84 ?
(1)
Ashu Pathak said:
7 years ago
Why not 3c1 * 8c2=84?
(1)
Shree said:
7 years ago
Thank you all for explaining the solution clearly.
Purvesh said:
7 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.
Murali said:
7 years ago
If these are the 3 takes then one ball should be black and the remaining balls can be taken as randomly, so
1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.
1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.
AkashSingh said:
7 years ago
1 Black + 1 White + 1 Red.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.
(1)
Suvam said:
7 years ago
@All.
My question is why red & white colour balls are taken together ( by saying non-black) why we don't take it separately?
I think answer should be;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
My question is why red & white colour balls are taken together ( by saying non-black) why we don't take it separately?
I think answer should be;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
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