Number of combinations with limited repetition

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The discussion centers on calculating the number of combinations for distributing n equivalent atoms among M distinct objects, with each object having a maximum limit of nlim atoms. When nlim is less than n, the problem diverges from the classical "Number of combinations with repetition." The user, Emile Maras, explores the complexity of the problem, noting that traditional summation methods become computationally intensive for large values of M and n. The example provided illustrates that for n=3 atoms and M=3 objects with a maximum of 2 atoms per object, there are 7 valid combinations.

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EmileMaras
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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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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 M^n. But because the atoms are identical,we should decrease this amount by dividing it by n!.
 
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.
 
EmileMaras said:
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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