# How many microstates have an assembly of Identical harmonic oscillators?

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We have an isolated assembly of N indistinguishable harmonic oscillators, each has energy $\epsilon_i=\hbar \omega/2 + n_i \hbar\omega$, where $n_i$ is a non-negative integer. If the total energy of the system is $E=N\hbar\omega/2 + M\hbar\omega$, ($N\gg 1$) then each micro-state must satisfy
$$\sum_{i=1}^{N}\epsilon_i =E\quad\Longrightarrow \quad \sum_{i=1}^N n_i=M$$
To determine micro-canonical entropy we need to know the number of possible ways for this relationship to be satisfied. I was trying to deduce a formula, or find some relationship of recurrence:

If $M=0$
$$n_i=0\quad\forall i\quad\Longrightarrow\quad\omega(E)=\dfrac{N!}{N!}=1,$$
If $M=1$
$$n_i=0\quad\forall i \not=j\;\;, \wedge, \;\; n_j=1\quad$$ $$\Longrightarrow\quad\omega(E)=\dfrac{N!}{(N-1)!}=N,$$
If $M=2$
$$\begin{cases}n_i=0\quad\forall i \not=j&, \wedge, \;\; n_j=2\\ n_i=0\quad\forall i \not=j_1,j_2&, \wedge, \;\; n_{j_1}=1= n_{j_2} \end{cases} \quad$$ $$\Longrightarrow\quad \begin{cases} \omega_1=\dfrac{N!}{(N-1)!}=N,\\ \omega_2=\dfrac{N!}{(N-2)!2!}=\dfrac{N(N-1)}{2},\\ \end{cases}\quad$$ $$\Longrightarrow\omega(E)=\omega_1+\omega=\dfrac{N(N+1)}{2}$$
In the same way, if $M=3$
$$\begin{cases} \omega_1=N\\ \omega_2=\frac{N!}{(N-2)!}=N(N-1)\\ \omega_3=\frac{N!}{(N-3)!(3!)}=\frac{N(N-1)(N-2)}{3!} \end{cases}\Longrightarrow\quad\omega(E)=\omega_1+\omega_2+\omega_3=\frac{N(N+1)(N+2)}{3!}$$
Then for a non-negative integer $M$,
$$\omega(E)=\frac{\displaystyle\prod_{i=0}^{M-1}(N+i) }{M!}\overset{?}{=}\frac{(N+M-1)!}{M!(N-1)!}= \begin{pmatrix} N+M-1\\N-1 \end{pmatrix}$$
But it is not clear to me, Is okay?, Is there another way to get this?, How I interpret this result?

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