# Deriving $\Omega$ for distinguishable particles

For atoms that are fixed in a crystal with the following assumptions:

1. There are quantised energy levels with energy $$E_i$$ for each atom.
2. Each state has a distinct energy $$E_i$$.

Such that..

$$N = \sum_i n_i$$ $$U = \sum_i E_i n_i$$

Where $$N$$ and $$U$$ are the total number of particles and total energy in the system respectively.

Why is $$\Omega$$ then..

$$\Omega = \frac{N!}{\prod_i n_i!}$$

I feel it must be derived from the combinations formula but I don't know how exactly. It's $$N$$ choose what and why?

• Are you sure this is for distinguishable particles? – nasu Apr 8 at 16:30

## 1 Answer

Consider that we randomly mix up all N particles, but consider fixed partitions that divide the particles into groups of $$n_i$$.

There are N! ways of mixing the particles up. But, if we mix up particles within a group, this does not correspond to a different macrostate.

So we overcount. By how much? For each group, there are $$n_i!$$ ways to mix the group's members, yielding a different microstate.

So we've overcounted by $$\prod_i n_i!$$. This explains the denominator.

• Just for extra-extra clarity, why does mixing the particles up within a group not correspond to a different macrostate? – sangstar Apr 14 at 16:58
• Because particles (or quanta here) are indistinguishable. – thedoctar Apr 15 at 6:42