Partition a box and put the same gas on both sides, at the same temperature and the same density. Then remove the partition. As the figure shows, the particles cross over and mix.
And what has changed? Temperature, density, pressure: nothing you can measure from outside has moved. Put the partition back and you have what you started with. Since it can be undone, the entropy cannot have gone up.
And yet, counted the way the piece on Boltzmann counted, it does go up. The room available has doubled, so \log W rises by N\log 2, which over both sides is 2Nk\log 2. That rise is written under the figure. This is the Gibbs paradox.
There is a case where the same calculation is right: when the two sides hold different gases. That is the lower box in the figure, drawn in two colours.
This one cannot be undone. Put the partition back into the mixture and all you get is half a mixture on either side. Separating them takes work, and that work is the rise in entropy. The rise really is 2Nk\log 2. It is called the entropy of mixing, and it can be measured.
Above and below, the particles do exactly the same thing. The only difference is whether they can be told apart once mixed. If a way of counting gives both the same answer, it is the counting that is wrong.
Let us find where it went wrong. When we counted states, we counted as one case each arrangement in which every single particle's place was settled. That is what we did in the first piece, calling it a different state depending on which particle was on the left.
That is counting as though the particles carried labels. The upper row of the figure is the labelled count: with three of them there are 6 arrangements. The lower row is the count with the labels off, where every arrangement collapses into the same one.
Particles of the same kind carry no labels. Swap two and there is no telling that anything was swapped. So counting with labels overcounts by a factor of N! for N particles. The answer came out right for two different gases because those really can be told apart.
The correction is straightforward. Divide out what was overcounted: divide W by N!.
Taking the logarithm subtracts \log N!, which Stirling's approximation turns into N\log N - N. As the figure shows, the naive Nk\log V becomes Nk\log \dfrac{V}{N}. The volume has taken a per-particle form.
That settles it. Double the system and both V and N double, so \dfrac{V}{N} does not move. The entropy then simply goes as N, which makes it properly a quantity that adds. Removing the partition between two halves of the same gas leaves \dfrac{V}{N} untouched, so the rise is 0.
Writing out the corrected entropy gives the formula in the figure. For an ideal gas it is known as the Sackur-Tetrode equation.
Volume and energy appear only in the per-particle forms \dfrac{V}{N} and \dfrac{E}{N}. So, as the last line says, doubling the system doubles the entropy exactly.
What is worth noticing is that Planck's constant h sits inside the formula. We are counting a classical gas, and yet the unit that slices phase space is borrowed from the quantum. Dividing by N! likewise comes, traced back far enough, from the quantum mechanics of identical particles. That is the end of the counting, for now. In the next chapter we stop fixing the energy and turn to systems whose temperature is fixed instead.