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.
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.
Ernest said:
1 decade ago
Mani is correct.
Since White balls and Red balls are *indistinguishable*, as the text does not say otherwise, choosing red ball 1 and red ball 2 is the same as choosing red ball 2 and red ball 3, and these are not new combinations.
Hence:
B,W,R
B,W,W
B,R,R
B,B,R
B,B,W
B,B,B
Are the only options.
Since White balls and Red balls are *indistinguishable*, as the text does not say otherwise, choosing red ball 1 and red ball 2 is the same as choosing red ball 2 and red ball 3, and these are not new combinations.
Hence:
B,W,R
B,W,W
B,R,R
B,B,R
B,B,W
B,B,B
Are the only options.
Pasindu said:
1 decade ago
According to the question can't we choose nine balls in these ways
please anyone explain this
(1-black,2-white),(2-black,1-white) ,(1- black,2-red), (2-black,1-red),(1-black,1-white,1red)and (3-black)
(3-black) is this correct or wrong if this wrong please tell me why is that soon
please anyone explain this
(1-black,2-white),(2-black,1-white) ,(1- black,2-red), (2-black,1-red),(1-black,1-white,1red)and (3-black)
(3-black) is this correct or wrong if this wrong please tell me why is that soon
Prateek said:
1 decade ago
Why 3c1*8c2 doesn't give the right answer...in 3c1 you select any one of the 3 black balls...then in in 8c2 you select any 2 balls out of remaining 8 balls which may or may not include other black balls.....in this way you will always have at least one ball.
Also 3c1 * 8c2 = 9c3... no idea why?
Also 3c1 * 8c2 = 9c3... no idea why?
Shiva said:
1 decade ago
It means all wt ball is identical&all red balls are identical&all black balls are identical
Now we can solve it by negetive method i.e
Total 3 balls are selected in 9c3 ways
at least 1 black ball = at most 0 black balls i.e 6c3 ways
So the ans is 9c3-6c3=64
Now we can solve it by negetive method i.e
Total 3 balls are selected in 9c3 ways
at least 1 black ball = at most 0 black balls i.e 6c3 ways
So the ans is 9c3-6c3=64
Vimal said:
1 decade ago
According to your point, the combination are 1 black and 2 white or 1 black and 2 red or 3 black. The other combination such as (1 black and 1w and 1r),(2b and 1w) or (2b and 1r) are missed here. So the answer is wrong for your assumption.
I hope you understood.
I hope you understood.
Sanjay said:
1 decade ago
There is totally 9 balls and we can draw any 3 balls = 9c3 ways. A.q. at least one black ball always included and there is 3 black ball if exclude all the 3 black ball there is remaining 6 the no.of selection is 6c3.
The total no.of selection = 9c3-6c3.
= 64.
The total no.of selection = 9c3-6c3.
= 64.
Nikolay said:
6 years ago
The total number of combinations can be 3x3x3 = 27 combinations of colours are possible.
But 1 out of 3 always have to be black so we exclude WWW and RRR, so this gives us 25 combinations - this has to be an answer. So, here option E is correct.
But 1 out of 3 always have to be black so we exclude WWW and RRR, so this gives us 25 combinations - this has to be an answer. So, here option E is correct.
Saireshwanth said:
1 decade ago
According to the question we can take (1black+1white+1red) , (1black+2white) , (1black+2red) , (2black+1white) , (3black) by this way we can take (3c1*2c1*4c1) + (3c1*2c2) + (3c1*4c2) + (3c2*2c1) + (3c2*4c1) by this way also the answer is 64.
RupamRD said:
9 years ago
2 White balls (W)
3 black balls (B)
4 red balls (R)
1W 1R 1B : (2c1 * 3c1 * 4c1)
1W 2B : (2c1 * 3c2)
1R 2B : (4c1 * 3c2)
2W 1B : (2c2 * 3c1)
2R 1B : (4c2 * 3c1)
3B + (3c3)
_______
64
_______
3 black balls (B)
4 red balls (R)
1W 1R 1B : (2c1 * 3c1 * 4c1)
1W 2B : (2c1 * 3c2)
1R 2B : (4c1 * 3c2)
2W 1B : (2c2 * 3c1)
2R 1B : (4c2 * 3c1)
3B + (3c3)
_______
64
_______
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