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Basics Concepts and Formulas in Permutations and Combinations


 
 
 
 
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sreekanth
2015-12-18 07:08:58 
I have a doubt the number of ways to wear 6 ring in 4 fingers is 4^6 but why not 6^4
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Anish
2016-01-23 15:01:57 
In the above problem,  you cannot think it like-
1st finger can be put in 6 rings
2nd finger can be put in 6 rings
...
4th finger can be put in 6 rings.
therefore total no. of ways=6^4

Because if you do so , you will consider two or more fingers can be put into a single ring, which is not possible.
so you cannot think the problem like that.
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sam
2015-12-22 12:56:50 
We have to think it in this way.
1st ring can go into any of the 4 fingers
2nd ring can go into any of the 4 fingers
3rd ring can go into any of the 4 fingers
4th ring can go into any of the 4 fingers
5th ring can go into any of the 4 fingers
6th ring can go into any of the 4 fingers

therefore answer is 4*4*4*4*4*4 = 4^6
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Hemant
2015-10-25 09:58:31 
How do I generate Permutation dynamically where number of position are dynamic and per position possible option is again dynamic?

e.g. I want to pick up 4 number  (here 4 number is dynamic) n1n2n3n4
and again for each number position i.e. 1 to 6 possible option will be dynamic.

1 st number could be = 1,2,3
2nd  number could be = 1
3rd number could be = 1,2
4th  number could be = 5,6,7

any algorithm get all those permutation?
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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?
(1) (0) Reply
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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adha
2015-11-20 04:25:06 
not to multiply all,  you have to sum all here
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Support, careerbless.com
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
(0) (0) Reply
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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