N single-particle states, n particles to place among them. How many distinct system states are there?
It depends entirely on what kind of particle you have: bosons allow any number per
state, fermions allow at most one (Pauli exclusion), and classical
particles are individually labeled, so swapping two of them counts as a new state even if the
occupation numbers look the same.
Bosons
Fermions
Classical
Multiplicity
Counting Formula
Why do the three counts differ so much?
A fermion state is a yes/no choice of which n of the N states are occupied — the same
counting problem as picking a committee, C(N,n). A boson state is a "how many balls in each of N
boxes" problem — stars and bars, C(n+N−1,n) — because nothing stops many
particles from crowding into the same state. A classical state additionally tracks which
labeled particle is where, so every one of the Nn independent choices counts separately
— including ones that look identical once you erase the labels.
That's why, for the same N and n, Fermions ≤ Bosons ≤ Classical always: erasing labels can
only ever merge classical states together, never split them apart, and exclusion only ever removes
boson states, never adds new ones.
This is exactly the same Ω that appears everywhere else in this series (Einstein solids,
paramagnets): a raw count of distinct microstates. S = kB ln Ω applies here just
as it does there — this sim just makes the individual states you're counting visible one by
one, instead of jumping straight to the total.
