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.
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.
Vikash said:
9 years ago
I agree @Sam.
If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.
If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.
Xyz said:
9 years ago
I agree @Vimal.
Balls are to be considered as identical and as one object.
Balls are to be considered as identical and as one object.
Bhuvi said:
9 years ago
It can be the balls are of same colours but different shades.
In my book, RD Sharma's objective mathematics it is done like 9C3 - 6c 3 which created doubt in my mind he took all the balls together which is not possible in case of identical balls. As there is only 1 way to select 2 balls from 20 identical balls.
In my book, RD Sharma's objective mathematics it is done like 9C3 - 6c 3 which created doubt in my mind he took all the balls together which is not possible in case of identical balls. As there is only 1 way to select 2 balls from 20 identical balls.
Naveen said:
8 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.
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.
Prince said:
8 years ago
6C2, 6C1..why we take 6 here?
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.
Sourav said:
8 years ago
Why not this?
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
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.
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