Number of combinations with limited repetition

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Discussion Overview

The discussion revolves around a combinatorial problem involving the distribution of equivalent atoms among distinct objects with a limit on the number of atoms each object can contain. Participants explore the implications of this limitation on the calculation of combinations, particularly as the number of objects and atoms increases while keeping the limit on atoms per object finite.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Emile Maras presents a combinatorial problem where n equivalent atoms are distributed among M distinct objects, each with a maximum limit (nlim) on the number of atoms they can contain.
  • Emile notes that when nlim is greater than n, the problem reduces to the classical "Number of combinations with repetition," but seeks to understand the case where nlim is less than n.
  • Emile expresses difficulty in calculating the number of combinations as M and n grow large, even with small values of nlim, and asks for a simpler analytical solution.
  • One participant suggests that for each atom, there are M choices, leading to a calculation of M^n, but acknowledges the need to adjust for identical atoms by dividing by n!.
  • Another participant challenges this approach by providing a specific example with n=3 atoms and M=3 objects, indicating that the proposed method does not yield the correct number of combinations.
  • The same participant reiterates the example to clarify the misunderstanding and notes that the initial answer provided was incorrect, leading to a non-integral value.

Areas of Agreement / Disagreement

Participants do not appear to reach consensus on the correct method for calculating the combinations under the specified constraints, with multiple interpretations and approaches being discussed.

Contextual Notes

The discussion highlights the complexity of combinatorial calculations when limits are imposed on the distribution of indistinguishable items, with participants grappling with the implications of these constraints on their proposed solutions.

Who May Find This Useful

Individuals interested in combinatorial mathematics, particularly those exploring problems involving distributions with constraints, may find this discussion relevant.

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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