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 2 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)
Bbbb said:
1 decade ago
@ Saireshwanth and Saurabh.
Could you explain how you get 64? Because I got the answer as 73.
Could you explain how you get 64? Because I got the answer as 73.
(1)
Jumenal said:
1 decade ago
How you taken 6 in combination?
(1)
Adith said:
1 year ago
Very simple, here we have to first find the 9c3 - 6c3.
(1)
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)
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)
Shree said:
8 years ago
Thank you all for explaining the solution clearly.
Ria kedia said:
8 years ago
@Krish. Same doubt.
Ankit said:
6 years ago
@All.
My solution was:
(considering balls are non-identical)
1. Case(1 Blackball), Possibilities: 1*6*5 = 30.
2. Case (2 Black balls), Possibilities: 1*1*6 = 6.
3. Case(3 Black balls), Possibilities: 1*1*1 = 1.
Total ways: 36
I am not getting the answer, please anyone help me to get the right answer.
My solution was:
(considering balls are non-identical)
1. Case(1 Blackball), Possibilities: 1*6*5 = 30.
2. Case (2 Black balls), Possibilities: 1*1*6 = 6.
3. Case(3 Black balls), Possibilities: 1*1*1 = 1.
Total ways: 36
I am not getting the answer, please anyone help me to get the right answer.
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.
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