I am having a hard time understanding the term used to calculate the $number$ of $microstates$ in the Boltzmann Distribution, I get that the term $$M=\dfrac{N!}{n_1!\times n_2!\times n_3!\times\cdots},$$ where:
i) $N!$ is the number of ways in which each atom can be arranged (under the assumption that each arrangement is unique).
ii) If $n_1$ is a macrostate, then $n_1!$ is the number of microstates that give the macrostate $n_1$; similarly $n_2$ and $n_3$ are the macrostates and $n_2!$ and $n_3!$ are the numbers of microstates that lead to the macrostates $n_2$ and $n_3$.
iii) Therefore the term $M$ tells us the number of distinct arrangements (distinct meaning that it doesn't lead to the same macrostate).
Is my understanding up to this point correct ?
Now the terms that confuse me are $g_1^{n_1}$ and $g_2^{n_2}$. What is degeneracy in layman terms? Why are we calculating that here?
Why are we calculating the number of sub -ontainers here?
https://i.stack.imgur.com/aNy5H.jpg
I am interested in this because I was fascinated by how a large collection of atoms is studied collectively and how it's used in various other calculations. My current level of knowledge is Calculus 1 & 2 and AP Physics; any effort to explain in this level is highly appreciated.
Best Answer
This answer is just summarizing the material on wikipedia.
Suppose there are $M$ energy levels, $E_1, E_2, \cdots, E_M$, and that we have $N$ distinguishable particles. For now, suppose that the energy levels are not degenerate. In other words, suppose there is one state associated with each energy level.
Then, we want to count the number of ways we can have $N_1$ particles in energy level 1, $N_2$ particles in energy level 2, etc.
There are \begin{equation} W = {N \choose N_1} = \frac{N!}{(N-N_1)!N!} \end{equation} ways $W$ to select $N_1$ particles to place in the first energy level. Then there are \begin{equation} W = {N-N_1 \choose N_2} = \frac{(N-N_1)!}{(N-N_1-N_2)! N_2!} \end{equation} ways to choose $N_2$ particles to go into the second energy level, from the remaining $N-N_1$ particles.
Multiplying these results together, we find there are \begin{equation} W= {N \choose N_1} {N-N_1 \choose N_2} = \frac{N!}{(N-N_1)!N!} \frac{(N-N_1)!}{(N-N_1-N_2)! N_2!} = \frac{N!}{N_1! N_2! (N-N_1-N_2)!} \end{equation} ways to have $N_1$ particles in the first energy level and $N_2$ particles in the second energy level.
Here, I will share a screenshot of the wikipedia article I linked above, which nicely shows the cancellation that occurs when computing the number of ways $W$ of having a given set of occupation numbers
The final step here is to realize that the last factor in the denominator in the second line is simply 1, since $N$ is equal to the sum of the occupation numbers in each state, so $N-\sum_{i=1}^M N_i = 0$ and then we use $0!=1$.
This leads to the result \begin{equation} W = N! \prod_{i=1}^M \frac{1}{N_i!} \end{equation} when ignoring degeneracies.
Now we include the possibility that energy level $i$ has a degeneracy of order $g_i$. This means that there are $g_i$ states that have the same energy $E_i$.
To account for this hiccup, consider what happens when we place the first of $N_i$ particles into the energy level $E_i$. There are $g_i$ choices for which state we place this particle into. Then we place the second particle, and again there are $g_i$ choices, so there are $g_i^2$ choices for placing particles 1 and 2. Carrying out this procedure for all $N_i$ particles, we find that there are $g_i^{N_i}$ ways of placing $N_i$ particles into this energy level.
Applying this correction to the previous result for $W$, we get the final result \begin{equation} W = N! \prod_{i=1}^M \frac{g_i^{N_i}}{N_i!} \end{equation}
Loosely speaking, the $1/N_i!$ factor is there to count the number of ways to place $N_i$ particles into an energy level $E_i$, and the $g_i^{N_i}$ factor is there to account for the number of ways to assign $N_i$ particles to the $g_i$ states within that energy level.