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 3 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?
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.
Sourav said:
8 years ago
Why not this?
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.
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.
Prince said:
8 years ago
6C2, 6C1..why we take 6 here?
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.
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.
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.
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.
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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