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A Random Logic Question on Probability

 
 
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A Random Logic Question on Probability
by wenhui zhang - Monday, January 18, 2021, 1:26 PM
 
A permutation is used to put unique objects into unique boxes. Combinations put unique objects into identical boxes. Stars and bars put identical objects into unique boxes. Then how do you find putting identical objects into identical boxes? I really don't know a method besides counting it out.


 
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Re: A Random Logic Question on Probability
by Duy Duong - Monday, January 18, 2021, 6:42 PM
 

well if they are all identical then there is only 1 way, is there not?

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Re: A Random Logic Question on Probability
by wenhui zhang - Monday, January 18, 2021, 7:29 PM
 

Well say I want to add three positive intergers to a total of 7. That means that I get (1,2,4), (1,1,5), (1,3,3), (2,2,3),  as my only choices, totaling 4. This is not stars and bars, as stars and bars would do order. This is just having an identical set of 7 ones and figuring out how to split it into ununique groups.

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Re: A Random Logic Question on Probability
by Charles Zhang - Tuesday, January 19, 2021, 8:52 AM
 
There is no way to count it unless you count it one-by-one, using casework.  The only case in which double-indistinct is avoidable is when you have 2 boxes.  Then it is (S+B)/2!


However, this does not generalize to n boxes.  If you have 3 boxes, you cannot do this because some cases cannot be permutated 3!=6 ways.

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Re: A Random Logic Question on Probability
by Dr. Kevin Wang - Saturday, January 23, 2021, 12:56 PM
 

The general case for this problem is pretty hard.

However, you should be able to derive a formula for 3 boxes.