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Number of combinations with limited repetition

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EmileMaras
#1
Dec12-13, 06:00 AM
P: 2
Hello

I have the following combinatoric problem :
I want to distribute n (equivalent) atoms among M distinct objects. Each object can contain from 0 to nlim atoms. How many combination do I have for this system?

If nlim>n, this problem actually corresponds to the classical "Number of combinations with repetition". But in my case nlim<n. In fact, I am interested in the limit of (lnΩ)/n (Ω beeing the number of combination) when M and n tend toward infinity (with n=a M where a is a constant) while nlim is finite (and actually rather small)..

I found a solution for that problem using some series of summations but it will be impossible to caculate as soon as M and n become large (even for M=100, n=300 and nmax=10, it took my laptop more than one hour to solve it).
Is there a simple analytical solution to this problem?

Thank you for your help.

Emile Maras
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Shyan
#2
Dec12-13, 06:37 AM
Shyan's Avatar
P: 859
For the first atom,there is M choices...for the second,again M choices...for the third,again M choices...and so on!
So there is always M possible choices and all that we should do is to multiply the number of choices for each of the atoms which becomes [itex] M^n [/itex]. But because the atoms are identical,we should decrease this amount by dividing it by [itex] n! [/itex].
EmileMaras
#3
Dec12-13, 08:32 AM
P: 2
I guess that it is not the correct answer. Maybe I did not state my problem properly, so I will just give an exemple.
Let's say I have n=3 atoms and M=3 object. An object can contain at max nmax=2 atoms. Then the possible combinations are 111, 012, 021, 102, 120, 201, 210 (where xyz gives the number of atom in each object) which corresponds to 7 combinations.

Shyan
#4
Dec12-13, 08:46 AM
Shyan's Avatar
P: 859
Number of combinations with limited repetition

Quote Quote by EmileMaras View Post
I guess that it is not the correct answer. Maybe I did not state my problem properly, so I will just give an exemple.
Let's say I have n=3 atoms and M=3 object. An object can contain at max nmax=2 atoms. Then the possible combinations are 111, 012, 021, 102, 120, 201, 210 (where xyz gives the number of atom in each object) which corresponds to 7 combinations.
Yeah,my answer is wrong.It even gives a non-integral value!
Anyway,Check here!


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