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 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways  = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{3}C_{3})  


= (45 + 18 + 1)  
= 64. 
Discussion:
78 comments Page 1 of 8.
Mani said:
1 decade ago
I have a doubt in this.
Its not mentioned in the question that all the balls are nonidentical. It is said '3 black balls', so all the 3 black balls are identical.
The required no. of ways = (1 black and 2 white) or (1 black and 2 red) or (1 black and 1 white and 1 red) or (2 black and 1 white) or (2 black and 1 red) or (3 black)
= 1+1+1+1+1+1
= 6ways.
Can anyone please confirm my answer?
Thank u :)
Its not mentioned in the question that all the balls are nonidentical. It is said '3 black balls', so all the 3 black balls are identical.
The required no. of ways = (1 black and 2 white) or (1 black and 2 red) or (1 black and 1 white and 1 red) or (2 black and 1 white) or (2 black and 1 red) or (3 black)
= 1+1+1+1+1+1
= 6ways.
Can anyone please confirm my answer?
Thank u :)
Rahul said:
1 decade ago
@ Mani: you are right in what you say but what you have answered is only half the question. what you have answered is in how many different combinations can the question requirements be satisfied.
(1 black and 2 white) = From those 9 balls you are specifically picking one black and two white balls here. The probability to do that would be 3/9 for the black or 1/3 then 2/8 or 1/4 for the first white ball (since a black is already removed) and 1/7 for the 2nd white ball. You will have to multiply those three numbers 1/3 x 1/4 x 1/7 to get the probability. By saying that the probability is 1 you are saying that there is a 100% guarantee that from those 9 balls you will always get 1 black and 2 white, which is obviously not the case. So, if you want to solve it this way I have done the first of the six combinations for you, you will need to do the other 5. This is a much longer method of course so you are better off doing the other method. :)
(1 black and 2 white) = From those 9 balls you are specifically picking one black and two white balls here. The probability to do that would be 3/9 for the black or 1/3 then 2/8 or 1/4 for the first white ball (since a black is already removed) and 1/7 for the 2nd white ball. You will have to multiply those three numbers 1/3 x 1/4 x 1/7 to get the probability. By saying that the probability is 1 you are saying that there is a 100% guarantee that from those 9 balls you will always get 1 black and 2 white, which is obviously not the case. So, if you want to solve it this way I have done the first of the six combinations for you, you will need to do the other 5. This is a much longer method of course so you are better off doing the other method. :)
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.
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 9c36c3=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 9c36c3=64
Pasindu said:
1 decade ago
According to the question can't we choose nine balls in these ways
please anyone explain this
(1black,2white),(2black,1white) ,(1 black,2red), (2black,1red),(1black,1white,1red)and (3black)
(3black) is this correct or wrong if this wrong please tell me why is that soon
please anyone explain this
(1black,2white),(2black,1white) ,(1 black,2red), (2black,1red),(1black,1white,1red)and (3black)
(3black) is this correct or wrong if this wrong please tell me why is that soon
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. :).
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
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 9c36c3=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 9c36c3=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.
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.
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