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 3 of 9.
Tejas said:
5 years ago
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.
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.
(1)
Pavel Sain said:
1 decade ago
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.
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.
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 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
Purvesh said:
7 years ago
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.
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.
Suvam said:
7 years ago
@All.
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;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
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;
(2c1x3c1x4c1)+(2c1x3c2x4c0)+(2c0x3c3x4c0).
= 31.
Naveen said:
9 years ago
@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.
So either you use the formula or make the individual combinations, the answer will be 6.
Murali said:
7 years ago
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.
1 8 7 =1 * 8 * 7 = 56. what is wrong in this, where did I make a mistake? Please tell me.
Sasi said:
1 decade ago
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).
In this question also why can't we do this as:
= (3C1 x 6C(6-2)) + (3C(3-2) x 6C1) + (3C3).
Viren Lakum said:
9 years ago
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.
(1 black and any 2) + (2 black and any 1) + ( 3 black)
(3C1 x 8C2) + (3C2 x 7C1) + (3C3) = 168 + 21 + 3 = 192.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers