$Combinatorics/counting: How many configurations are possible for m differenct objects in n boxes of unlimited occupany (m<n)$

Question

Hello! The question is pretty much entirely in the title. I wasn't sure if I could just use the fundmental principle of counting here with m^n, as when I look up related formulae in statistical mechanics I get a different result.
Just to be clear:
Say there are 5 distinguishable objects and 20 distinguishable boxes. I can put up to 5 objects in a box. I am interested in how many different configurations, or states, are possible under these cirumstances.
What would change if the objects were indistinguishable?

I'm interested in a little supporting justification, just so I understand the answer. Thank you, geniuses!

Discrete Mathematics Combinatorics

Halpneeded

Report

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

The problem becomes easier if you fix $n$ and do an induction on $m$. Suppose the number of boxes is fixed. Let $\alpha(m)$ be the number of ways you can put $m$ different object inside the $n$ different boxes. Then $\alpha(1)=n$. Suppose we can place $m$ objects inside the boxes in $\alpha(m)$ different ways. To place the $(m+1)^{th}$ object, we have $n$ options, so
\[\alpha(m+1)=n \alpha(m).\]
By a simple induction argument, we get
\[\alpha(m)=n^m,\]
which is the number of ways you can place $m$ different objects in $n$ different boxes.

Nirenberg

Nirenberg

Please leave a comment if you need any clarifications.

$Combinatorics/counting: How many configurations are possible for m differenct objects in n boxes of unlimited occupany (m<n)$

Answer

Related Questions

Search