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:
82 comments Page 8 of 9.
Chandu said:
1 decade ago
There is 9 balls is there of different colours from this 9 balls we select 3 so that we select: 9C3 ways and we donot considered 3 black balls,
So that we have 6 balls in this 6 balls we considered 3 balls so that we select : 6C3 ways and finally by subtracting from total selection - without containing black balls we get resulting answer: 10C3 - 6C3 = 64.
So that we have 6 balls in this 6 balls we considered 3 balls so that we select : 6C3 ways and finally by subtracting from total selection - without containing black balls we get resulting answer: 10C3 - 6C3 = 64.
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.
Garry said:
1 decade ago
I believe Mani is correct. Consider this: From 2 red balls, you must choose 1. How many ways are there of doing this ?
From two balls, you can choose 1, 2C1 or 2. But there is only one possible outcome: {1 red ball}. Since the two red balls are identical, there is actually only 1 way to choose, not 2.
So there are actually 6 ways of choosing as Mani said, not 64.
From two balls, you can choose 1, 2C1 or 2. But there is only one possible outcome: {1 red ball}. Since the two red balls are identical, there is actually only 1 way to choose, not 2.
So there are actually 6 ways of choosing as Mani said, not 64.
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.
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
Biswajit said:
1 decade ago
In the question at least 1 black ball should be in that box.
So 3 black ball should not be selected.
ans- 63
So 3 black ball should not be selected.
ans- 63
Tamsha said:
1 decade ago
No 3 black is correct because dey have said there should b atleast 1 black ball. So you can take more then 1 but not less then 1. So eventually 3 is also correct. :).
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
Kranthi said:
1 decade ago
Consider no black ball is dran
there are totally 9 balls in which 3 balls can dran in 9c3 ways
we need to pick the 3 ball which are not black
we can do it in 6c3 ways
No of ways we can select is 9c3-6c3=64
there are totally 9 balls in which 3 balls can dran in 9c3 ways
we need to pick the 3 ball which are not black
we can do it in 6c3 ways
No of ways we can select is 9c3-6c3=64
Saurabh said:
1 decade ago
Mani acoording to you,
If you take (1black+2red) or (1black+2 white) or (1black+1 red+1 white) or (2 black+ 1red) or (2black+1 white)
Then the answer will be
(3C1*4c2)+(3c1*2c2)+3c1*4c1*2c1)+(3c2*4c1)+(3c2*2c1)=64.
If you take (1black+2red) or (1black+2 white) or (1black+1 red+1 white) or (2 black+ 1red) or (2black+1 white)
Then the answer will be
(3C1*4c2)+(3c1*2c2)+3c1*4c1*2c1)+(3c2*4c1)+(3c2*2c1)=64.
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