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Basics Concepts and Formulas in Permutations and Combinations



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Shiksha
2015-01-12 18:17:16 
Hi,
I have a question, if we have 6 white balls, 3 blue balls and 2 Red balls and we have to pick a selection 4?
a) how many combinations it will be?
b) if always 2 white have to be selected from 4 then how many combinations possible?
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Shraya
2015-08-17 17:09:13 
a). Your head question and part a both depict the same. Picking any four irrespective of the color.
So, 6 + 3 + 2 = 11.

11C4 = @@\dfrac{11!}{4!×(11-4)!}=11×10×3=330@@

b) The 2nd part i think you didn't give a clarity. Out of 4 if at least two must be white.

then,
picking 4 whites out of 6 whites = 6C4
picking 3 whites with 1 blue or red = 6C3 × 3C1 × 2C1
picking 2 whites with 2 blue = 6C2 × 3C2
picking 2 whites with 2 red = 6C2 × 2C2
picking 2 whites with 1 red 1 blue = 6C2 × 3C1 × 2C1

Now multiply all.
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2015-08-17 16:47:51 
This question has been moved to Discussion Board. Click here to view.
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Pankaj Sharma
2015-08-07 11:52:03 
as coming to your que 1
1) we can select any 4 balls out of (6B,3B,2R) so total combinations are 11C4
2)now in the second part exactly 2 balls are white so combinations are 6C2*5C2
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Guglani
2015-08-17 15:53:00 
i got the point i.e 6C2, here you select 2 balls from all six white balls, 
but u can have all the white balls also, i.e 4.
i think there would be 6C2*9C2 ??
isn't it !!!
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Asif
2015-08-23 06:47:38 
when we select 6C2, for the next selection we will not consider those 6 white balls...therefore next selection will be from remaining 5 balls out of 11 balls..ie 5C2..ans might be 6C2*5C2
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Gaurav Dobriyal
2015-01-23 15:40:35 
6  3   2
w  b   r
2  1   1  = No. of ways arrangement is done.
1  2   1
1  1   2

Thus 6c2+3c1+2c1+6c1+3c2+2c1+6c1+3c1+2c2 = 15+3+2+6+3+2+6+3+1=41

if always 2 white then
6c2+3c1+2c1 = 15+3+2=20
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Rajesh
2015-01-25 04:45:07 
But shouldn't it be 11c4 for (a).

the combinations could also be

6  3  2
w  b  r
0  3  1
4  0  0
2  2  0

and so on.. Nowhere it is mentioned that all colours need to be picked.So it can be done by 11C4 rite..?11c4=330

and for (b) shouldn't it be 6c2*3c1*2c1 + 6c2* 3c2 + 6c2*2c2 

Kindly correct me if I am wrong...
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Gaurav Dobriyal
2015-01-24 17:18:52 
sorry the signs went wrong this one is correct

6c2*3c1*2c1+6c1*3c2*2c1+6c1*3c1*2c2 = 90+36+18 = 144

and

if always 2 white then

6c2*3c1*2c1=15*3*2=90
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vinodraina
2014-03-17 08:42:03 
please write down inequality trick
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