I have read that in combinations with repetitions, the number of ways of selecting r objects out of n is: $\binom{n+r-1}{r}$ I have also read that this is the number of positive solutions for the linear equation of the form: $x_1 + x_2 + x_3 + x_4 +\cdots + x_r = n\qquad(E)$
We can model equation $(E)$ as a case where we fill up $r$ boxes with $n$ digits from $0$ to $9$. It can be graphically represented as: $xx\ldots x|xx\ldots x|xx\ldots x|\ldots|xx\ldots x|xx\ldots x$ where $|$ represents a partition between two boxes (hence we have $(r-1)$ partitions for $r$ boxes).
The number of solutions of $(E)$ is the number of ways in which we can arrange these $(r-1)$ $|$'s and $n$ $x$'s. This is equal to $\frac{(n+r-1)!}{n!(r-1)!}$. This is easy because we have to arrange $(n+r-1)$ symbols where n are one type and $(r-1)$ are of another. This expression can, however, be written as $\binom{n+r-1}{r}$ .
Here is my doubt: Is this is a coincidence? If no, then what is the combinatorial argument for it? How do we interpret such questions of permutations in terms of combinations?
I really hope that I was able to convey my doubt. If no, then please tell where I need to provide more information.