quantum mechanics - Degeneracy of the isotropic harmonic oscillator

The degeneracy for an $p$-dimensional quantum harmonic oscillator is given by [1] as

$$g(n,p) = \frac{(n+p-1)!}{n!(p-1)!}$$

The $g$ is the number of degenerate states. Where of course we have that $n = n_1 + n_2 + \dots + n_p$ and that the eigenvalues of the energy are given by

$$E_n = E_{n_1,n_2,\dots,n_p} = \hbar \omega\left(n + \frac{p}{2}\right)$$

How can I derive this formula? In the paper as reference I just didn't understand what he used to show that relation. If this question has an answer that is too big for this format then I'll accept a hint to show that, for $p = 3$ we have that $$g_n = \frac{1}{2}(n+1)(n+2) \, .$$ Can this can be deduced in a counting way?

Answer

The formula can be written as $$g= \binom{n+p-1}{p-1}$$ it corresponds to the number of weak compositions of the integer $n$ into $p$ integers. It is typically derived using the method of stars and bars:

You want to find the number of ways to write $n= n_1 + \cdots + n_p$ with $n_j \in \mathbb{N}_0.$ In order to find this, you imagine to have $n$ stars ($\star$) and $p-1$ bars ($|$). Each composition then corresponds to a way of placing the $p-1$ bars between the $n$ stars. The number $n_j$ corresponds then to the number of stars in the $j$-th

For example ($p=3, n=6)$:

$$\star \star | \star \star \star | \star \quad \Rightarrow \quad n_1=2,n_2=3, n_3=1$$ $$\star| \star \star \star \star \star| \quad \Rightarrow \quad n_1=1,n_2=5, n_3=0.$$

Now it is well known that choosing the position of $p-1$ bars among the $n+(p-1)$ objects (stars and bars) corresponds to the binomial coefficient given above.