We keep the box from the previous piece. Each of the N particles is either on the left or on the right. One such settlement is what we call one state. Since we tell the particles apart, there are 2^{N} states in all.
Saying that n of them are on the left does not name a single state. It says that there are W = \binom{N}{n} states answering to that description. The figure takes N = 6 with n = 2: every one of them has two on the left, and every one is a different state.
Once we grant that every state happens as often as every other, a larger W means a more likely description. That grant is the principle of equal a priori probability, and the next chapter takes it head on. For now we take it as given and look only at how W is counted.
W gets out of hand at once. Even the middle of N = 100 passes 10^{29}, and at N = 10^{23} the number of digits is itself astronomical. There is no use carrying a number you cannot write down.
So we look at \log W instead. Where W climbs exponentially with N, \log W climbs in plain proportion to it. The figure has W on the left and \log W on the right, the same thing drawn two ways.
Factorials cannot be differentiated as they stand, so we replace them by \log N! \simeq N\log N - N. This is Stirling's approximation, and the larger N is the better it fits. Now \log W is something you can write by hand and later differentiate with respect to n.
Adding every state gives \displaystyle\sum_{n} \binom{N}{n} = 2^{N}. The largest single term, the one bar where the two sides are even, stands at about 2^{N} divided by \sqrt{N}.
So the sum and its largest term differ only by a factor of \sqrt{N}. Take logarithms and the difference is \log\sqrt{N}, which vanishes beside N. The coloured bar in the figure is that largest term.
We may therefore replace the whole sum by its largest term. This is the method of the largest term, and statistical mechanics uses it from here on. Instead of adding everything up, we hunt for the highest point.
Position and momentum are continuous, so they cannot be counted the way left and right were. Take a plane with position q across and momentum p up. It is called phase space. One state is one point of that plane, and points cannot be counted.
So we rule the plane into cells and call one cell one state. The size of a cell is Planck's constant h. A product of q and p has the dimensions of action, and so does h, so dividing an area by h leaves a bare number. The grid in the figure is those cells.
With f degrees of freedom the phase space has 2f dimensions, and the state count is its volume divided by h^{f}. That the cell size is h comes out of quantum mechanics, but here it is enough to say that counting something continuous needs a unit.
Take N particles of an ideal gas in a box and count the states with energy up to E. The condition that the kinetic energies sum to no more than E is the inside of a sphere of radius \sqrt{E} in a 3N dimensional space. Its volume makes W go as E raised to the power \dfrac{3N}{2}.
That exponent \dfrac{3N}{2} is of order 10^{23}. Raise E by a mere 1\% and W grows by 1.01 raised to that power. The curve in the figure stands like a wall even though the exponent drawn there is far smaller.
The speed of that growth is what becomes temperature in the next chapter: the derivative of \log W with respect to E turns out to be \dfrac{1}{T}. The counting tools we need are all here.