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I recently got introduced to the Statistical Mechanics, particularly, the Statistical Interpretation of Entropy and am utterly confused regarding the following problem:

Imagine a box with two identical compartments. Then a crude specification of the position of a molecule of a gas will be whether it's located in the left or the right compartment of the box. Any molecule is equally likely to be in either of the compartments.

Assume that there are $N$ 'identical' molecules.

Consider the macrostate with $n_1$ = $k$ and $n_2$ = $N-k$. Here, $n_1$ refers to the number of molecules in the left compartment and $n_2$ to the number of molecules in the right compartment.

Then the number of microstates corresponding to this macrostate should be 1 (since all the particles are identical).

But, in all the resources I've referred to, I found the number of microstate(s) corresponding to this macrostate to be $\frac{N!}{n_1! n_2!}$.

Kindly throw some light onto this conceptual flaw.

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Then the number of microstates corresponding to this macrostate should be 1 (since all the particles are identical).

The particles may be identical (have the same intrinsic physical properties), but that does not in any way mean that they are not distinct. It is most natural to assume the particles are distinct (they can be placed to different places, labeled and tracked).

The microstate of the system is not given by the occupation numbers of the departments, but by a set of locations for each distinct particle. Equivalently, it is given by the set of particles in the first department.

The number of such sets $\Omega$ is easily calculated. Imagine a process where we create a particular set by placing $k$ particles from the initial $N$-membered set into the first department. The number of distinct ways to do that is

$$ W = N(N-1)(N-2)...(N-(k-1)) = \frac{N!}{(N-k)!}. $$

However, when making one particular set, $k!$ distinct ways are possible (differing in the order in which the particles were placed). The number of possible sets is thus lower by a factor of $k!$:

$$ \Omega = \frac{N!}{k!(N-k)!}. $$

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  • $\begingroup$ What an irony of fate! We both answered at the same time; deduced the right thing but yours is upvoted! Don't know what wrong I've written in my ans:P Let me give you +1, also:D $\endgroup$ – user36790 Sep 27 '15 at 13:03
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It's a simple permutation problem.

Suppose there are $N$ distinguishable particles. Let $n_L$ be the number of particles distributed in the left & $n_R$ distributed at the right compartment.

First of all, you are to select a particle for the first place of $N$ places. How many ways can it be done? It can be done in $N$ ways; now the second place can be filled by $(N-1)$ ways. Similarly, by extrapolation, we can find the number of ways the $N^\text{th}$ place can be filled is $(N-(N-1))$ ways. By rule of multiplication, all these can be done simultaneously in $N!$ ways.

Now, suppose you've $n_L$ distinguishable particles ; then they can be arranged in $n_L$ places by $n_L!$ ways; similarly $n_R$ particles can be arranged among $n_R$ places by $n_R!$ ways.

But all we know that $N$ particles are indistinguishable. Let the total number of arrangements be $x$. For each arrangement, we could arrange $n_L$ particles & $n_R$ particles among them by $n_L!$ & $n_R!$ ways if they were distinguishable. So, total number of arrangements is $x\cdot n_L!\cdot n_R!$ ways. But this is equal to $N!$ ways, So, by equating them, we get the number of ways the indistinguishable particles can be arranged i.e. $x$ viz.$$x= \frac{N!}{n_L!\cdot n_R!}.$$

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