Correct Boltzmann Counting

Write down the partition function of a gas of $N$ identical molecules and you face a subtle trap. The natural move is to take the single-particle partition function $z_1$ and raise it to the $N$ -th power — one factor per particle. But that quietly assumes the particles wear labels, and identical molecules do not: swapping molecule 3 with molecule 7 produces the same physical microstate, not a new one. The labelled phase-space integral counts every genuine configuration $N!$ times over. The cure is a single, deceptively humble factor: divide by $N!$ . This is correct Boltzmann counting, and without it classical statistical mechanics is quietly broken. The symptom of the error is a failure of extensivity. Entropy, free energy, and volume ought to double when you double the system at fixed density — a bucket of gas plus an identical bucket has twice the entropy. The uncorrected entropy does not: it carries a spurious $k_BN\ln N$ term that makes it grow faster than the system size. The most famous face of this defect is the Gibbs paradox — the absurd prediction that removing a partition between two halves of the same gas increases the entropy. Dividing by $N!$ turns the offending $V^N$ into approximately $(V/N)^N$ , restores extensivity to the last decimal, and makes the paradox vanish. Gibbs introduced the $1/N!$ on purely thermodynamic grounds, decades before quantum mechanics existed. Its deeper origin is quantum: identical particles are fundamentally indistinguishable, and in the dilute, non-degenerate limit where each single-particle state is almost never doubly occupied, the exact Bose-Einstein or Fermi-Dirac counting collapses to exactly $z_1^N/N!$ . The factorial is the last fingerprint of quantum indistinguishability surviving into the classical world — and it is the reason the Sackur-Tetrode entropy of argon matches the calorimeter to a fraction of a percent.

Key takeaways

A gas of $N$ identical particles has partition function $z_N=z_1^N/N!$ , not $z_1^N$ : the labelled phase-space integral overcounts each physical microstate by the $N!$ permutations of identical particles.
Dividing by $N!$ — "correct Boltzmann counting" — restores the extensivity of $S$ , $F$ , and $\mu$ and resolves the Gibbs paradox of mixing identical gases.
Without it, the entropy carries a spurious $k_BN\ln N$ term and grows superextensively (off by $2Nk_B\ln 2$ when you double the system).
The $1/N!$ turns the non-extensive $V^N$ into the extensive $(V/N)^N$ , and the chemical potential $\mu=k_BT\ln(n\lambda^3)$ becomes intensive as it must be.
It is the unique prefactor that fixes extensivity: no constant $c^N$ can do the job, only $N!$ (via Stirling's $N\ln N-N$ ).
The factor is the classical, dilute limit ( $n\lambda^3\ll1$ ) of quantum indistinguishability; near degeneracy it fails and full Bose/Fermi statistics are required.
Key idea: identical particles are indistinguishable, so the labelled classical measure overcounts each microstate by $N!$ ; dividing $z_N$ by $N!$ is exactly what makes $S$ , $F$ , and $\mu$ extensive and resolves the Gibbs paradox.

The core insight

Identical particles are indistinguishable, so the labelled classical phase-space measure counts each physical microstate $N!$ times; dividing $z_N$ by $N!$ restores the extensivity of the entropy, free energy, and chemical potential and dissolves the Gibbs paradox — and this factor is the classical shadow of quantum indistinguishability.

Where this lives today

Correct Boltzmann counting is quietly load-bearing across physics and chemistry. Its most direct experimental test is the Sackur-Tetrode entropy: the absolute molar entropies of the noble gases, computed from $S=Nk_B[\ln(V/(N\lambda^3))+5/2]$ , match the values obtained by integrating measured heat capacities up from absolute zero to a fraction of a percent — a routine but striking confirmation that the $1/N!$ is physically real. The intensive chemical potential $\mu=k_BT\ln(n\lambda^3)$ that only the factor produces is the foundation of the law of mass action, and therefore of every equilibrium constant in chemistry. In astrophysics, the Saha ionization equation — which sets the degree of ionization in stellar atmospheres and predicts spectral classes — balances chemical potentials that each carry their own species' $1/N_i!$ and thermal wavelength. And in the theory of dilute solutions, the entropy of mixing $-k_B\sum_i N_i\ln x_i$ , which survives correct counting only because it distinguishes solute from solvent, yields van 't Hoff's law of osmotic pressure. Wherever an entropy, a chemical potential, or an equilibrium constant is computed for identical particles, the $1/N!$ is doing silent work.

The physical picture

Start with the simplest possible caricature: two identical coins. If you secretly tag them A and B, you count "A-heads, B-tails" and "A-tails, B-heads" as two distinct arrangements. But if the coins are genuinely identical, you cannot tell those apart — there is one arrangement, "one heads and one tails." The classical phase-space integral always carries hidden labels (it integrates over particle 1, particle 2, and so on separately), so it systematically overcounts. For $N$ particles spread over many more accessible cells than there are particles, almost every configuration has all particles in distinct cells, and the overcount is a uniform $N!$ — which is exactly why a single constant divisor cures it. Think of the $1/N!$ as a "permutation tax" that converts a sum over labelled arrangements into a sum over genuine, physically distinct configurations. The tax is uniform in the dilute regime because double occupancy is negligible; every microstate is overcounted by the same $N!$ , so one factor pulled out front fixes them all at once.

Figure

Two identical particles in labelled boxes: swapping labels gives the SAME physical microstate. The labelled integral counts it twice (N! = 2 times for N=2); dividing by N! keeps one copy.

The $1/N!$ is the precise place where quantum indistinguishability bleeds into classical mechanics. Even after $\hbar$ has been sent to zero everywhere else in the calculation, the symmetry of the wavefunction under particle exchange leaves behind exactly $1/N!$ (alongside the $1/h^{3N}$ that fixes the absolute size of a phase-space cell). The diagnostic that tells you the factor is doing real work is extensivity: the free energy must satisfy $F(\lambda N,\lambda V,T)=\lambda F(N,V,T)$ . The punchline figure below shows what happens if you forget it — the entropy per particle, which should be a flat horizontal line, instead climbs without bound as $\ln N$ .

Figure

Punchline: entropy per particle S/N versus system size N at fixed density. WITH 1/N! it is flat (extensive, blue). WITHOUT it, S/N grows like ln N (superextensive, red) — the Gibbs defect, unbounded and unphysical.

The core equations

The whole subject rests on one modification to the classical partition function: an explicit division by $N!$ for a system of $N$ identical, freely exchanging particles. This is the object from which all thermodynamics is then computed.

Z_N = \frac{1}{N!}\,\frac{1}{h^{3N}}\int d^{3N}q\;d^{3N}p\; e^{-\beta H(q,p)} \qquad (1)

where: $Z_N$ — the $N$ -particle canonical partition function; $N!$ — the Gibbs counting factor that removes the permutation overcount; $h^{3N}$ — the phase-space cell size making $Z_N$ dimensionless (a second quantum remnant); $\beta=1/(k_BT)$ ; $H$ — the classical Hamiltonian. Reading it: both $1/N!$ and $1/h^{3N}$ are corrections that the pre-quantum theory could not justify but that the quantum classical-limit supplies exactly. For non-interacting particles the Hamiltonian separates, and the multi-particle partition function factorizes into single-particle pieces — divided by the counting factor.

Z_N=\frac{z_1^{\,N}}{N!},\qquad z_1=\frac{V}{\lambda^3},\qquad \lambda=\frac{h}{\sqrt{2\pi m k_BT}} \qquad (2)

where: $z_1$ — the single-particle partition function; $V$ — volume; $\lambda$ — the thermal de Broglie wavelength; $m$ — particle mass. Reading it: $z_1=V/\lambda^3$ counts how many thermal-wavelength-sized cells fit in the box, and $z_1/N=V/(N\lambda^3)$ is intensive — the phase-space occupancy per particle. Taking $F=-k_BT\ln Z_N$ and applying Stirling's approximation to $\ln N!$ produces the extensive free energy — the payoff of the whole exercise.

F=-k_BT\ln Z_N=-Nk_BT\left[\ln\!\left(\frac{V}{N\lambda^3}\right)+1\right] \qquad (3)

where: $V/N$ — the volume per particle. Reading it: because $F$ depends on $N$ and $V$ only through the intensive ratio $V/N$ (times an overall $N$ ), it scales linearly with system size — extensivity restored. The crucial $-\ln N!\approx -N\ln N+N$ is what converts the offending $\ln V$ into $\ln(V/N)$ . Differentiating $F$ with respect to temperature gives the Sackur-Tetrode absolute entropy, and with respect to $N$ gives the chemical potential — both intensive-per-particle, both finite in the thermodynamic limit only because of the $1/N!$ .

S=Nk_B\left[\ln\!\left(\frac{V}{N\lambda^3}\right)+\frac52\right],\qquad \mu=k_BT\ln(n\lambda^3) \qquad (4)

where: $n=N/V$ — the number density; the $5/2$ splits as $3/2$ (from the energy, $-\partial F/\partial T$ ) plus $1$ (from the counting factor). Reading it: $S$ and $\mu$ depend only on the intensive $n\lambda^3$ , so doubling the system doubles $S$ and leaves $\mu$ unchanged — exactly what a well-behaved entropy and chemical potential must do.

Derivation AWhy 1/N! restores extensivity — and why nothing else can

The goal is to prove that dividing $z_1^N$ by $N!$ makes the entropy exactly extensive, $S(\lambda N,\lambda V,\lambda E)=\lambda S(N,V,E)$ , and that no simpler prefactor — in particular no constant raised to the $N$ -th power — can do the same. The strategy is to compute the entropy with and without the factor, apply Stirling to $\ln N!$ , and test homogeneity of degree one under a joint rescaling of $N$ and $V$ at fixed intensive variables.

Step 1. With the factor, $F=-k_BT[N\ln(V/\lambda^3)-\ln N!]$ . Apply Stirling $\ln N!\approx N\ln N-N$ : the bracket becomes $N\ln(V/\lambda^3)-N\ln N+N=N[\ln(V/(N\lambda^3))+1]$ , so $F=-Nk_BT[\ln(V/(N\lambda^3))+1]$ , equation (3).

Step 2. Differentiate: $S=-\partial F/\partial T=Nk_B[\ln(V/(N\lambda^3))+5/2]$ , where the extra $1$ over the naive $3/2$ comes from the temperature dependence of $\lambda^{-3}\propto T^{3/2}$ combined with the $+1$ from Stirling.

Step 3. Rescale $N\to\lambda N$ , $V\to\lambda V$ at fixed $T$ (so $\lambda_{\text{th}}$ unchanged): $S(\lambda N,\lambda V)=\lambda Nk_B[\ln(\lambda V/(\lambda N\lambda_{\text{th}}^3))+5/2]=\lambda Nk_B[\ln(V/(N\lambda_{\text{th}}^3))+5/2]=\lambda S(N,V)$ . The scale factor cancels inside the logarithm because $V$ and $N$ both scale — extensivity holds exactly. Why a constant $c^N$ fails. Suppose instead you divided $z_1^N$ by some $c^N$ . Then $F=-k_BT[N\ln(V/\lambda^3)-N\ln c]$ , and the entropy still contains a bare $N\ln V$ term (not $N\ln(V/N)$ ). Rescaling gives a leftover $\lambda N k_B\ln\lambda$ defect — the same superextensive term as with no factor at all. Only $\ln N!\sim N\ln N$ , which supplies the missing $-N\ln N$ , can convert $\ln V$ into $\ln(V/N)$ .

Step 4. Without any factor, $S_{\text{uncorr}}=Nk_B[\ln(V/\lambda^3)+3/2]$ , and $S_{\text{uncorr}}(\lambda N,\lambda V)-\lambda S_{\text{uncorr}}(N,V)=\lambda Nk_B\ln\lambda$ . Doubling ( $\lambda=2$ ) leaves a surplus of $2Nk_B\ln2$ — the Gibbs defect, unbounded as the system grows. The extensivity identity $S(\lambda N,\lambda V)=\lambda S(N,V)$ holds with the factor; without it the superextensive defect is $\lambda Nk_B\ln\lambda$ (i.e. $2Nk_B\ln2$ on doubling); and no constant $1/c^N$ prefactor can restore extensivity.

Go deeper · Derivation B · 1 minEmergence of 1/N! from quantum indistinguishability

Derivation BEmergence of 1/N! from quantum indistinguishability

The goal is to show that the phenomenological $1/N!$ is not an ad hoc fix but the exact classical (dilute, non-degenerate) limit of quantum statistics, where each single-particle state is almost never multiply occupied. The strategy is to write the canonical partition function as a sum over occupation numbers of single-particle levels, take the dilute limit where all occupancies are $0$ or $1$ , and count how many labelled terms map to each unlabelled configuration.

Step 1. In quantum mechanics a many-body state is specified by the occupation numbers $\{n_k\}$ of single-particle levels $\varepsilon_k$ , with no particle labels at all: $Z_N=\sum_{\{n_k\},\,\sum n_k=N}\prod_k e^{-\beta\varepsilon_k n_k}$ . The (anti)symmetrization is entirely encoded in which $\{n_k\}$ are allowed and in the absence of labels.

Step 2. In the dilute limit $n\lambda^3\ll1$ the mean occupancy $\langle n_k\rangle\ll1$ , so terms with any $n_k\ge2$ are negligible and the Bose/Fermi distinction disappears — only $n_k\in\{0,1\}$ contribute.

Step 3. Compare to the labelled product $(\sum_k e^{-\beta\varepsilon_k})^N=z_1^N$ . Expanding it, each unlabelled configuration with all occupancies $\le1$ appears exactly $N!/\prod_k n_k!=N!$ times (since all $n_k!=1$ ). Dividing by $N!$ collapses these $N!$ labelled permutations to the one physical configuration: $Z_N=z_1^N/N!$ . Why the factorial and not something else. The overcount is $N!/\prod_k n_k!$ ; only in the dilute limit, where every $n_k!=1$ , does it reduce to a clean $N!$ independent of the configuration. This is precisely why a single constant divisor works classically but fails near degeneracy, where doubly-occupied states ( $n_k=2$ , contributing $n_k!=2$ ) make the overcount configuration-dependent.

Step 4. The same result follows from expanding the ideal-gas grand potential $\ln\Xi=\pm\sum_k\ln(1\pm e^{-\beta(\varepsilon_k-\mu)})$ to first order in the fugacity $z=e^{\beta\mu}\ll1$ : the leading term reproduces $z_1^N/N!$ , with the $\pm$ (boson/fermion) corrections appearing only at the next order in $z$ . The dilute-limit overcount ratio (distinguishable assignments)/(indistinguishable) $=N!$ , as illustrated numerically for several $(M,N)$ pairs above; the reduction $Z_N\to z_1^N/N!$ is the standard classical limit of quantum statistics.

Limiting cases

The $1/N!$ is a specific correction with a specific domain — get the domain wrong and you introduce new errors.

Case	Prescription	Meaning
Dilute classical gas, $n\lambda^3\ll1$	$Z_N=z_1^N/N!$	The simple factorial is exact to leading order; gives Sackur-Tetrode.
Distinguishable particles (lattice sites)	$Z_N=z_1^N$ (no $1/N!$ )	Atoms pinned to distinct sites are labelled by position; permutations ARE distinct states.
Mixture of species	$Z_N=\prod_i z_{1,i}^{N_i}/N_i!$	One factorial per species; mixing different gases gives a real $\Delta S=-k_B\sum_i N_i\ln x_i$ .
Quantum degenerate, $n\lambda^3\gtrsim1$	full Bose/Fermi statistics	Multiple occupancy matters; the constant $1/N!$ over- or under-counts.

Common trapThe 1/N! is a universal correction you always apply

The factorial is not universal. Three mistakes recur. First, do not divide by $N!$ for distinguishable particles: atoms fixed on distinct lattice sites (an Einstein solid) are labelled by their positions, so their permutations are genuinely different states — dividing by $N!$ there would break extensivity, not restore it. Second, for a mixture, divide by $\prod_i N_i!$ (one factorial per species), never by $(\sum_i N_i)!$ — otherwise you would spuriously cancel the real entropy of mixing distinct gases. Third, the constant $1/N!$ is only valid in the dilute limit $n\lambda^3\ll1$ ; near quantum degeneracy it fails entirely and you must use full Bose-Einstein or Fermi-Dirac statistics. The rule is precise: one factorial per indistinguishable, freely-exchanging species, in the dilute regime.

Worked examples

Example 1 — Extensivity of the entropy, with and without the factor (enumerate, apply, boundary number). Take one mole of argon ( $N=N_A=6.022\times10^{23}$ ) at $298\,\mathrm K$ , $1\,\mathrm{atm}$ , and ask what happens to the entropy when you place an identical mole beside it and remove the partition (doubling $N$ and $V$ at fixed density).

Step 1. Enumerate the two formulas. With the factor, $S=Nk_B[\ln(V/(N\lambda^3))+5/2]$ ; without it, $S_{\text{uncorr}}=Nk_B[\ln(V/\lambda^3)+3/2]$ .

Step 2. Apply the doubling. With the factor, $V/N$ is unchanged, so $S(2N,2V)=2Nk_B[\ln(V/(N\lambda^3))+5/2]=2S(N,V)$ — exactly additive. Without the factor, $S_{\text{uncorr}}(2N,2V)=2Nk_B[\ln(2V/\lambda^3)+3/2]=2S_{\text{uncorr}}(N,V)+2Nk_B\ln2$ .

Step 3. Boundary number. The spurious surplus per mole is $2N_Ak_B\ln2=2R\ln2=2(8.314)(0.693)=\mathbf{11.5\,J\,K^{-1}}$ — a large, unmistakable, and entirely unphysical entropy that appears from merely relabelling identical atoms. Correct Boltzmann counting removes it exactly. This $2R\ln2$ is the Gibbs paradox in a single number.

Example 2 — The overcount factor for a small system (an actual integer). Consider $3$ identical particles distributed one each among $100$ available single-particle states (dilute, so no double occupancy).

Step 1. Count the labelled arrangements: particle 1 has $100$ choices, particle 2 has $99$ , particle 3 has $98$ , giving $100\times99\times98=970200$ labelled assignments.

Step 2. Count the physically distinct configurations: choosing which $3$ of the $100$ states are occupied, $\binom{100}{3}=161700$ .

Step 3. Boundary number. The ratio is $970200/161700=\mathbf{6}=3!$ — exactly the number of permutations of $3$ particles. The labelled count overcounts by $N!=6$ , and dividing by $3!$ recovers the correct number of microstates. This is the combinatorial heart of the $1/N!$ , verified here for a concrete case.

Origins · 1 minindistinguishable particles and how this was discovered

indistinguishable particles

statistical-mechanics concept · Gibbs 1902, quantum justification 1920s

Correct Boltzmann counting is the prescription $Z_N=z_1^N/N!$ for indistinguishable particles. Despite the name, the $1/N!$ was introduced not by Boltzmann but by Josiah Willard Gibbs in his 1902 Elementary Principles in Statistical Mechanics, on the thermodynamic grounds that entropy must be extensive and that mixing identical gases must produce no entropy change. Gibbs recognized the paradox that now bears his name and prescribed the factorial as its cure, purely phenomenologically — quantum mechanics did not yet exist. The deeper justification came only in the 1920s, when the classical (dilute) limit of the newly-formulated Bose-Einstein and Fermi-Dirac statistics was shown to reproduce exactly $z_1^N/N!$ . The factor is thus a piece of quantum indistinguishability that Gibbs, remarkably, deduced from thermodynamic consistency a generation early.

History

1875

— Gibbs identifies the mixing paradox for identical gases.

1902

— Gibbs prescribes division by N! in Elementary Principles in Statistical Mechanics.

1911

— Sackur, and 1912 Tetrode, obtain the absolute entropy of a monatomic gas using the factor.

1924

— Bose and Einstein formulate the statistics whose dilute limit yields 1/N!.

1926

— Fermi and Dirac give the antisymmetric statistics, whose classical limit also reproduces 1/N!.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.5–1.6, §3.5 · Kardar, Statistical Physics of Particles, Ch. 4 · Reif, Fundamentals of Statistical and Thermal Physics, §7.3 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, §3.5, §7.3 · Landau & Lifshitz, Statistical Physics Part 1, §31, §41. Key numerical benchmarks: (i) $F=-Nk_BT[\ln(V/(N\lambda^3))+1]$ is extensive after Stirling; (ii) $S(\lambda N,\lambda V)=\lambda S(N,V)$ with the factor, and the superextensive defect $\lambda Nk_B\ln\lambda$ (i.e. $2Nk_B\ln2$ on doubling) without it; (iii) no constant $c^N$ prefactor restores extensivity; (iv) the dilute-limit overcount ratio equals $N!$ ; (v) argon's molar entropy $\approx154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ at $298\,\mathrm K$ , $1\,\mathrm{atm}$ . Constants used throughout: $h=6.62607\times10^{-34}\,\mathrm{J\,s}$ , $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Classical Ideal Gas · Thermal de Broglie Wavelength · Gibbs Paradox Leads to: Sackur-Tetrode Equation · Classical Limit of Quantum Statistics · Classical Ideal Gas

Why is it called "Boltzmann" counting if Gibbs introduced the factor?

The name is a historical accident. The method of counting microstates by distributing particles among energy cells goes back to Boltzmann's combinatorial derivation of the distribution law, so "Boltzmann counting" refers to that general procedure. The specific $1/N!$ correction that makes it correct for identical particles was Gibbs's contribution in 1902, addressing the paradox that also bears his name. So "correct Boltzmann counting" means "Boltzmann's counting method, with Gibbs's factorial fix." Neither man had quantum mechanics; the deeper justification of the factor as the classical limit of indistinguishable-particle statistics came two decades later.

How exactly does dividing by N! fix the Gibbs paradox?

The paradox is that the uncorrected entropy predicts $\Delta S>0$ when you remove a partition between two halves of the same gas at equal density — physically absurd, since nothing observable changes. The uncorrected entropy is $Nk_B[\ln(V/\lambda^3)+3/2]$ , which depends on $\ln V$ separately from $N$ ; combining two subsystems then produces a spurious $\ln$ -of-volume gain. Dividing by $N!$ turns $\ln V$ into $\ln(V/N)$ , so the entropy depends only on the intensive density $N/V$ . When you merge two identical gases at the same density, $V/N$ is unchanged, so the total entropy is exactly the sum of the parts and $\Delta S=0$ . For genuinely different gases, the real entropy of mixing $-k_B\sum_i N_i\ln x_i$ survives, as it should.

Is the 1/N! an approximation, or is it exact?

It is exact in the dilute, non-degenerate regime $n\lambda^3\ll1$ , and an approximation (indeed a failure) near quantum degeneracy. In the dilute limit essentially every single-particle state has occupancy $0$ or $1$ , so every physical configuration is overcounted by precisely $N!$ in the labelled sum, and the constant divisor is exact to leading order. When $n\lambda^3$ approaches $1$ , states begin to be multiply occupied; the overcount then becomes configuration-dependent ( $N!/\prod_k n_k!$ ), and a single constant $N!$ no longer captures it. There you must abandon correct Boltzmann counting and use the full Bose-Einstein or Fermi-Dirac statistics, which count indistinguishable states correctly at any occupancy.

Why don't we divide by N! for a crystal of N atoms on lattice sites?

Because those atoms are distinguishable — each is permanently labelled by the lattice site it occupies. You can point to "the atom at site (3,5,2)" unambiguously, and swapping two atoms between sites produces a genuinely different microstate. There is no permutation overcounting to remove, so $Z_N=z_1^N$ with no factorial. In fact, dividing a lattice solid's partition function by $N!$ would break its extensivity rather than restore it. The $1/N!$ applies only to particles free to exchange positions, like the molecules of a gas or liquid, where you cannot say which molecule is which.

Where does the extra "1" in the Sackur-Tetrode 5/2 come from?

The $5/2$ decomposes as $3/2+1$ . The $3/2$ is the energy contribution: differentiating the free energy with respect to temperature picks up $\tfrac32 Nk_B$ from the $T^{3/2}$ dependence of $\lambda^{-3}$ (equivalently, $U=\tfrac32 Nk_BT$ gives $\tfrac32 Nk_B$ in the entropy). The extra $+1$ comes directly from the counting factor: Stirling's $\ln N!\approx N\ln N-N$ contributes a $+N$ term that, after differentiation and per-particle bookkeeping, adds $1$ to the bracket. So the famous $5/2$ is a fingerprint of correct Boltzmann counting — without the $1/N!$ you would get an entropy with $3/2$ (and a non-extensive $\ln V$ instead of $\ln(V/N)$ ).

Does the h^{3N} play the same role as the 1/N!?

They are companion quantum remnants that fix different things. The $1/h^{3N}$ sets the absolute size of a phase-space cell: the classical integral $\int d^{3N}q\,d^{3N}p$ has units of (action) $^{3N}$ , and dividing by $h^{3N}$ makes $Z_N$ dimensionless and fixes the zero of entropy. The $1/N!$ fixes the permutation overcount from indistinguishability. Neither can be justified within purely classical mechanics; both emerge together when you take the classical limit of the quantum partition function. Only with both factors does the classical $Z_N$ reproduce the correct, extensive, absolute thermodynamics — and match experiment, as the Sackur-Tetrode entropy shows.