How many microstates have an assembly of Identical harmonic oscillators?

Question

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?