Define $f(n,m)$ as returning the amount of ways to throw $n$ distinguishable balls into $m$ bins, and $g(n,m)$ as returning the amount of ways to throw $n$ indistinguishable balls into $m$ bins. I am wondering whether anyone could conceptually justify why $\frac {f(n,m-1)} {f(n,m)} = \frac {g(n,m-1)} {g(n,m)} $ for any integers $n,m >1$, without just counting the numerators and denominators and using algebraic manipulation to show equivalence. So far, I have tried to think of these functions recursively: seeing combinatorially how $f(n,m)= f(n-1,m)+ f(n,m-1)$ and $g(n,m) = g(n-1,m-1) + g(n,m-1)$- but I haven't gotten anywhere conceptual with this (There may be an inductive proof with this, but this isn't the type of justification I want here unless it is extremely simple and intuitive).
This problem is of my own construction- the motivation for asking about it was to understand the underlying phenomena in the problem here better.
Just so we can close this question, the two ratios you ask about are not equal.
It is easy to prove that $g(n,m)=m^n$, because each of the $n$ balls must be placed into one of the $m$ bins. This means that $$ \frac{g(n,m-1)}{g(n,m)}=\left(\frac{m-1}{m}\right)^n $$ However, using stars and bars, we can show $f(n,m)=\binom{m+n-1}{n}$. Therefore, $$ \frac{f(n,m-1)}{f(n,m)}=\frac{\binom{m+n-2}{n}}{\binom{m+n-1}{n}}=\frac{m-1}{m+n-1} $$ Clearly, these two expressions are not equal, for example when $m=2$ and $n=1$.