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.
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. :)
Miles said:
9 years ago
Consider the case where only one black ball is chosen. It could be chosen first second or third so that's 3 possibilities. As for the other two balls, they could be red or white. It could be red then white, white then red, white then white, or red then red, regardless of when the black ball is chosen.
So that's 4 possibilities. So, with only one black ball, that would be 12 possibilities (brw, bwr, brr, bww, rbw, wbr, rbr, wbw, rwb, wrb, rrb, wwb) but the solutions says there are 45 possibilities for this condition. Am I missing something or what?
So that's 4 possibilities. So, with only one black ball, that would be 12 possibilities (brw, bwr, brr, bww, rbw, wbr, rbr, wbw, rwb, wrb, rrb, wwb) but the solutions says there are 45 possibilities for this condition. Am I missing something or what?
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 :)
Mukesh said:
10 years ago
Here you can draw one black ball from x number of black balls in only one way, similarly 2 red balls from x numbers of red balls also in one way for xC2.
So, (3 Black) + (2 Black AND (1 Red OR 1 White)) + (1 Black AND ((2 white OR 2 Red OR (1 White AND 1 Red))).
> 1 + 1 * 2 + 1 * (1+1+1).
> 1 + 2 + 3.
> 6.
Which I believe should be answer, any corrections or clarification most appreciated.
So, (3 Black) + (2 Black AND (1 Red OR 1 White)) + (1 Black AND ((2 white OR 2 Red OR (1 White AND 1 Red))).
> 1 + 1 * 2 + 1 * (1+1+1).
> 1 + 2 + 3.
> 6.
Which I believe should be answer, any corrections or clarification most appreciated.
Amrish said:
7 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.
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.
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.
Mitali said:
8 years ago
If we consider that the balls are not identical, then one black ball for sure can be selected in 3C1 ways, after that, you can select any 2 out of the remaining 8 in 8C2, which will include black also. Can someone explain why this 3C1 x 8C2 also equals the total number of ways of selecting 3 balls out of 9 (i.e. 9C3)?
Vish said:
2 years ago
We can also do it this way,
Atleast one black ball = total balls  No black ball
= 9c3  6c3.
9c3= 3 balls selected from the total no of balls,
6c3  3 balls taken from the total no of balls without any black which is 6(2 white + 3 red).
By calculating this, we will get the answer.
Atleast one black ball = total balls  No black ball
= 9c3  6c3.
9c3= 3 balls selected from the total no of balls,
6c3  3 balls taken from the total no of balls without any black which is 6(2 white + 3 red).
By calculating this, we will get the answer.
(12)
Bhuvi said:
8 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.
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