# Number of microstates compatible with two boxes

From my notes I have: From one point of view there are many more microstates compatible with the LHS than the RHS, in fact the relation between the number of microstates is $$\frac{n_B}{n_A}=\left(\frac{1}{2}\right)^N$$

I cannot understand why we have the $1/2$ to the power $N$. I would have thought that it would be $$\frac{n_B}{n_A}=\frac{1}{2}$$ as there is half the room. Could someone please explain the reasoning behind the power of $N$ please.

$$\frac{n_{B}}{n_{A}}=\frac{\int_{V/2}d^{3N}q d^{3N}p}{\int_{V}d^{3N}q d^{3N}p}=\frac{\int_{V/2}d^{3N}q}{\int_{V}d^{3N}q}=\frac{\int_{V/2}d^{3}q_1 \int_{V/2}d^3 q_{2}\dots }{\int_{V}d^{3}q_1 \int_{V}d^3 q_{2}\dots}=\left( V/2 \right)^N/V^N=\frac{1}{2^N}$$ where $N$ is the number of particles and $V$ is the volume of the box.
This may be the argument: you have $N$ particles, and for each one you can put it on the left side or on the right side. Each of these choices, for each particle, leads to a different microstate. There are $2^N$ possible choices you can make for how to distribute the $N$ particles between the left and right halves of the box (assuming the particles are distinguishable), and thus $2^N$ microstates, but only one of these microstates has the particles all on the right side. So the relation is $n_B/n_A = 1/2^N$. This assumes that diagram A is meant to represent the macrostate containing all possible microstates, even those in which the particles do happen to all wind up on the right side.
If that sounds somewhat unsatisfying, here's a slightly better argument. Suppose the large box is divided into $2M$ little squares, where $M$ is very large. The number of ways to place the particles into these $2M$ squares is $$n_A = \binom{2M - 1 + N}{N} = \frac{(2M - 1 + N)!}{N!(2M - 1)!}$$ now assuming the particles are indistinguishable (though it turns out not to matter, since distinguishable particles just have an extra $N!$ in both this expression and the following one). But the number of ways to place the particles into the $M$ squares on the right half of the box is $$n_B = \binom{M - 1 + N}{N} = \frac{(M - 1 + N)!}{N!(M - 1)!}$$ The ratio of these is $$\frac{n_B}{n_A} = \frac{(M - 1 + N)!}{N!(M - 1)!}\frac{N!(2M - 1)!}{(2M - 1 + N)!} = \frac{M(M+1)\cdots(2M - 1)}{(M+N)(M+N+1)\cdots(2M-1+N)}$$ and if you take the logarithm, you get \begin{align} \ln\frac{(M - 1 + N)!}{(M - 1)!}&\frac{(2M - 1)!}{(2M - 1 + N)!} \\ \approx\ &(M-1+N)\ln(M-1+N) - (M-1+N) \\ &- (M-1)\ln(M-1) + (M-1) \\ &+ (2M-1)\ln(2M-1) - (2M-1) \\ &- (2M-1+N)\ln(2M-1+N) + (2M-1+N) \\ \approx\ &(M+N)\ln(M+N) - M\ln M + 2M\ln(2M) - (2M+N)\ln(2M+N) \\ =\ &M\biggl(\ln\frac{4}{2+r} + (1+r)\ln\frac{1+r}{2+r}\biggr) \end{align} where $r = \frac{N}{M}$. In the limit of extremely large $M\gg N$, so that $r\to 0$, this can be approximated as $$\ln\frac{n_B}{n_A} = -Mr\ln 2 + \mathcal{O}(r^2) = -N\ln 2 + \mathcal{O}(r^2)$$ which leads directly to $$\frac{n_B}{n_A} = \frac{1}{2^N}$$ But again, this is based on the assumption that state A represents the macrostate consisting of all possible microstates, even those where the particles are all on one side.