When four indistinguishable particles are distributed in two identical compartments the total number of microstates is?

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Consider a system of three identical and distinguishable non-interacting particles and three available nondegenerate single particle energy levels having energies 0, ɛ and 2ɛ. The system is in contact with a heat bath of temperature T K. A total energy of 4ɛ is shared by these three particles. How many ways can the particles be distributed?

Compartment-wise distributions of a system of particles are known as macrostates of the system. There are five macrostates observed corresponding to a system of four particles . In general, for a system of n particles, there are exactly n+1 macrostates (for the system of n particles to be distributed in two identical compartments). These macrostates are actually the following distribution of particles:

d1=(0,n)

d2=(1,n−1)

d3=(2,n−2)

…

dk=(k−1,n−k+1)

…

dn+1=(n,0)

It can be observed that a macrostate may have many corresponding microstates. The distribution d1=(0,4) has one microstate {{ }{a,b,c,d}}, while the distribution d2=(1,3) has four, d3 has six , d4 has four and d5 has one.

The total number of microstates for the system of four particles is, therefore, 1+4+6+4+1=16=24. In general, total number of microstates for the system of n-particles in ‘two compartment’ composition is 2n.

First of all, if there are $p^N$ possible states, this means that the particles and the compartments are distinguishable. For indistinguishable particles (and distinguisable compartments, which is probably reasonable physically speaking) the number of possible states, or configurations, has to be divided. For instance the configuration $(n_1,n_2,\dots,n_p)$ of indistinguishable particles corresponds to $$\frac{N!}{n_1!n_2!\dots n_p!}$$ configurations of distinguishable particles. Note that we have of course $$\sum_{{\substack{n_1,n_2,\dots,n_p\\n_1+n_2+\cdots+n_p=N}}} \frac{N!}{n_1!n_2!\dots n_p!}=p^N.$$

Let us start with the answer to the question "how many configurations have no compartment empty ?"

With indistinguishable particles, we place $N$ particles in $p$ compartments, with at least one particle in each. To count this number, we use the bullet representation of the integer $N$ like this $$\bullet\bullet\bullet\bullet\cdots\bullet$$ there are $N$ bullets. We now place the compartment separations, denoted as $|$. One possible configuration with $N=10$ into $p=5$ compartments is $$\bullet|\bullet\bullet\bullet|\bullet\bullet\bullet|\bullet|\bullet\bullet$$ which is a reprensentation of $(1,3,3,1,2)$. We have to place $p-1$ divisors at any $N-1$ positions. Therefore the number of possible divisions is $\binom{N-1}{p-1}$ (basically, we have counted the number of partition of the integer $N$ into sums of non-zero integers when order matters). There are therefore $$C_0(N,p)=\binom{N-1}{p-1}\quad\text{(indistinguishable particles)}$$ configurations with no empty compartments.

In the case of distinguishable particles, this actually more difficult, as @ndrv pointed out. We have $p^N$ possibilities, but we have to substract configurations with one empty compartment. We can pick up any of the $p$ compartment as empty and substract $p\times (p-1)^N$, but then we have substracted the configurations with two empty compartments two times. So must add $\binom{p}{2}(p-2)^N$. But by doing this we have also added twice the configurations with three empty compartments, so must add them back. This goes on until we have added or substracted the configuration with $p-1$ empty compartments, since there are no further configurations. As a conclusion, we have $$C_0(N,p)=\sum_{j=0}^{p-1}(-1)^j\binom pj(p-j)^N \quad\text{(distinguishable particles)}$$ This number is related to the Stirling number of second kind by $C_0(N,p)=p!\genfrac\{\}{0pt}{}Np$.

We can answer to the question "How many configurations have exactly one empty compartment ?" The answer is, in both cases $$pC_0(N,p-1)$$ because there remains only the choice of the empty compartment among the $p$ possible compartments.

We can even answer to the question "How many configurations have exactly $k$ empty compartments ?" From the same reasoning, we find $$\binom pkC_0(N,p-k)$$

If we now want to allow one or more empty compartments, we can sum over $k$: $$C(N,p)=\sum_{k=1}^{p-1}\binom pkC_0(N,p-k).$$ but we can also remark that $$C(N,p)=p^N-C_0(N,p).$$