Aptitude - Permutation and Combination - Discussion

Discussion :: Permutation and Combination - General Questions (Q.No.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?

[A]. 32
[B]. 48
[C]. 64
[D]. 96
[E]. None of these

Answer: Option C


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)
= 3 x 6 x 5 + 3 x 2 x 6 + 1
2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

Mani said: (Oct 4, 2010)  
I have a doubt in this.

Its not mentioned in the question that all the balls are non-identical. 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: (Oct 28, 2010)  
@ 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. :)

Saurabh said: (Jun 17, 2011)  
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


Kranthi said: (Jul 27, 2011)  
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

Pasindu said: (Oct 12, 2011)  
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

Tamsha said: (Nov 17, 2011)  
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: (Dec 21, 2011)  
In the question at least 1 black ball should be in that box.
So 3 black ball should not be selected.
ans- 63

Shiva said: (Feb 20, 2012)  
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

Ernest said: (May 30, 2012)  
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.


Are the only options.

Garry said: (Jun 17, 2012)  
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.

Saireshwanth said: (Mar 15, 2013)  
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.

Chandu said: (Jul 19, 2013)  
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.

Sanjay said: (Aug 29, 2013)  
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.

Sandeep said: (Oct 17, 2013)  
Hi, at least one black ball is to be included in the draw.

So 3bc1*(2w+2b+4r)c2.

Why above answer is not correct? can any one explain that.

Vimal said: (Nov 5, 2013)  
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.

Rohit said: (Dec 21, 2013)  
Guys why 'C' is taken in this. I could not understand please somebody can explain me in simple way.

Anupam Patra said: (Jul 3, 2014)  
Why can't I do like this.

One black ball is compulsory, So 9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final by 7 balls, that makes it 1*8*7=56.

Pradeep said: (Jul 16, 2014)  

9-1=8 balls fill the remaining 2 slots, the next slot by 8 balls and then the final is also by 8 balls, that makes it 1*8*8=64.

Got it!

Prateek said: (Jul 30, 2014)  
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?

Sam said: (Sep 25, 2014)  
If balls are IDENTICAL and ORDER does NOT matter - 6 ways.

If balls are IDENTICAL and ORDER matters - 19 ways.

If balls are NOT IDENTICAL - 64 ways.

Krish said: (Dec 7, 2014)  
3c1*8c2 = 84?

Why not? Atleast one black ball is there right? And from remainder we take 2 balls. What is wrong in this?

Sasi said: (Dec 18, 2014)  
I have a doubt. In last question you have given nC(n-r). Why This conditions is not suitable for this?

In this question also why can't we do this as:

= (3C1 x 6C(6-2)) + (3C(3-2) x 6C1) + (3C3).

Bbbb said: (Dec 28, 2014)  
@ Saireshwanth and Saurabh.

Could you explain how you get 64? Because I got the answer as 73.

Mukesh said: (Feb 4, 2015)  
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.

Jumenal said: (Feb 25, 2015)  
How you taken 6 in combination?

Tanya said: (May 7, 2015)  
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.

Pavel Sain said: (Jun 19, 2015)  
May be we can take (1B+1W+1R) OR (1B+2W) OR (1B+2R) OR (2B+1W) OR (2B+1R) OR all (3B).

So required no = (3C1*2C1*4C1)+(3C1*2C2)+(3C1*4C2)+(3C2*2C1)+(3C2*4C1)+3C3.

= (3*2*4)+(3*1)+(3*6)+(3*2)+(3*4)+1.

= 24+3+18+6+12+1 = 64.

Sundar said: (Jul 10, 2015)  
We should not take 3 black balls. Then why final declaration 3 ball added?

Prasoon Garg said: (Oct 26, 2015)  
I agree with @Mani as it is not mentioned in the question that all the balls are non-identical.

Sanjay said: (Oct 27, 2015)  
Why not 3C1*8c2?

Miles said: (Nov 29, 2015)  
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?

Himanshu Kulkarni said: (May 20, 2016)  
One black ball from 3 is 3C1 and any 2 balls from remaining 8 balls
i.e. 8C2.

So, why 3C1 x 8C2 does not match the solution?

Revathy said: (Jul 8, 2016)  
I think Pavel Sain's answer is correct.

Naveen said: (Jul 20, 2016)  
@Pavel Sain: Since the ball of the same colour are identical, you should do ((3C1/3!) * (2C1/2!) * (4C1/4!)) etc.

So either you use the formula or make the individual combinations, the answer will be 6.

Apluv said: (Jul 31, 2016)  
Answer is 64. If the colors are different then balls are not identical.

Gauss said: (Aug 13, 2016)  
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?

Anil said: (Sep 13, 2016)  
Can anybody tell me why the answer is not 3C1 x 8C2 = 84?

Mitali said: (Sep 14, 2016)  
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)?

Viren Lakum said: (Oct 11, 2016)  
There are 9 balls in box, and as given one of them has to be black so we get,
(1 black and any 2) + (2 black and any 1) + ( 3 black)
(3C1 x 8C2) + (3C2 x 7C1) + (3C3) = 168 + 21 + 3 = 192.

Rupamrd said: (Oct 15, 2016)  
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)

Chiru said: (Nov 7, 2016)  
It's simple.
The combination of all balls - neglecting combination of remaining balls (except at least one ball) = at least one black.
9c3 - 6c3 = 64.

Vikash said: (Feb 2, 2017)  
I agree @Sam.

If the balls are identical and having the same color then we can have only one way of selecting 1 black from 3 black.

Xyz said: (Feb 15, 2017)  
I agree @Vimal.

Balls are to be considered as identical and as one object.

Bhuvi said: (Mar 3, 2017)  
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.

Naveen said: (Jun 15, 2017)  
I think it should be clearly specified whether balls of particular colour are identical or not.

If identical, then answer would be 6.

Fuji said: (Jul 11, 2017)  
I selected 1 ball from 3 black ball. That reduces it to 2. So I get 2+2+4 balls of white, black and red respectively. So I get 8c2 ways. Only 59.

Am I correct? Please Explain.

Prince said: (Jul 20, 2017)  
6C2, 6C1..why we take 6 here?

Kalai said: (Jul 24, 2017)  
Hi @Fuji.

You take 3 balls at a time and that's all. You don't keep it aside. Every time you take is a new combination. I hope you are understanding it.

Sourav said: (Aug 21, 2017)  
Why not this?

3c1* (2c1*4c1) +3c2 (2c1+4c1) +3c3.

Aki47 said: (Aug 25, 2017)  
Why not this?

1 black ball = 3 ways.
Remaining are 4+2+2 = 8 balls.
Remaining 2 can be chooses in 8 x 7 = 56 ways.
No. of combinations thus = 3 x 56 = 168 ways.

Sanjay Verma said: (Sep 19, 2017)  
3C1*6C2 which means they have merged red and white balls?

Am I right here?

Amrish said: (Sep 24, 2017)  

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.

Ria Kedia said: (Nov 6, 2017)  
@Krish. Same doubt.

Supratik said: (Nov 12, 2017)  
Why not 3C1 x 8C2=84 ?

Ashu Pathak said: (May 4, 2018)  
Why not 3c1 * 8c2=84?

Shree said: (May 21, 2018)  
Thank you all for explaining the solution clearly.

Purvesh said: (Sep 7, 2018)  
Why is the difference between this 3c1 * 8c2=84.

And this;
1 black ball = 3 ways.
Remaining are 4+2+2 = 8 balls.
Remaining 2 can be chooses in 8 x 7 = 56 ways.
No of combinations thus = 3 x 56 = 168 ways.

Murali said: (Sep 27, 2018)  
If these are the 3 takes then one ball should be black and the remaining balls can be taken as randomly, so

1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.

Akashsingh said: (Oct 20, 2018)  
1 Black + 1 White + 1 Red.
2 Black + 1 White.
1 Black + 2 White.
2 Black + 1 Red.
2 Red + 1 Black.
3 Black.

Suvam said: (Nov 5, 2018)  

My question is why red & white colour balls are taken together ( by saying non-black) why we don't take it separately?

I think answer should be;
= 31.

Ankit said: (May 30, 2019)  

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.

Debrupa Sen said: (Jul 1, 2019)  
Here the question is black balls are selected from 3 black balls. In solution, it is seen 6 black balls. How it is possible?

Ankit said: (Jul 16, 2019)  
If I do like this;

(2C1*3C1*4C1)+(2C1*3C2*4C0)+(2C0*3C2*4C1)+(2C0*3C3*4C0). = (24+6+12+1)=43 WAYS.

Is this way right? Please tell me.

Nikolay said: (Jul 31, 2019)  
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.

Monika said: (Sep 24, 2019)  
We can say;

(Atleast one = Total - Zero).
So it's, 9C3- 3C0 * 6C3 = 64.

Anand said: (Dec 1, 2019)  
How (3C3)=1. Can anyone explain this, please?

Nilesh Sonune said: (Apr 8, 2020)  
Can it be like, 9C3-(2C2*4C1+2C1*4C2)= 68?

Tejas said: (Jun 19, 2020)  
Why the following method is wrong? Please tell me.

He have total 9 balls out of which 3 to be selected.

out of 3, one must be black.. out of 3 black we select 1. Now we have 8 balls remaining and a total of 2 places hence 8c2=28.

Stige said: (Aug 10, 2020)  
(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

How to write this and (3C1 x 6C2) + (3C2 x 6C1)?

Please, anyone, explain.

Shubham said: (Sep 17, 2020)  
Total ways - no blackball = Atleast one black ball.
i.e,, 9C3-6C3.

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