An exact formula for counting solutions of the Frobenius equation summed to 8

Question

Let us consider a Frobenius equation with: $$ x_1+\dots+x_n=8, \tag{1}$$ where $$(x_1, \dots, x_n)$$ must consist of non-negative integers, i.e. $$ x_j \in \mathbb{N} $$ as Natural numbers.

Here is my question: Is there a general formula for Eq.(1) counting all the possible solutions $$(x_1, \dots, x_n)$$
for given the positive integer $n \in \mathbb{Z}^+$? This should be related to the Partition, but I am not sure the exact forms are known? Say, can we find the total number of possible solutions as a function $f(n)$, and what is $$ f(n)=? $$

I am interested in finding $$ f(20)=? $$ $$ f(36)=? $$

p.s. Sorry if this question is too simple for number theorists. But please provide me answer and Refs if you already know the answer. Many thanks!

p.s.2. A more advanced generalized version of question is asked in https://mathoverflow.net/questions/352331/request-for-an-exact-formula-related-to-a-partition-in-number-theory

Answer 1

The number of solutions will be \begin{eqnarray*} [x^8]: (1+x+x^2+\cdots)^n = [x^8]: \frac{1}{(1-x)^n}. \end{eqnarray*} Now use \begin{eqnarray*} [x^m]: \frac{1}{(1-x)^n}= \binom{m+n-1}{n-1}. \end{eqnarray*}