I'm trying to organize the formulas for permutations and combinations.
Let's say we sample $r$ objects from an urn which contains $n$ objects, $m$ of which are duplicates. So if the objects in the urn are $2$ red balls, $3$ blue balls, $4$ yellow balls, and $5$ green balls, then the number of balls is $n = 14$, the number of color duplicates is $m = 10$ (thus the number of colors is $n - m = 4$), and the sample size is $r = 3$. One formula appears to require the individual color tallies; I'll call this arbitrarily ordered sequence $u$ (in the ball example, $u$ contains $2$, $3$, $4$, and $5$, in whatever order).
It seems that the three fundamental categorical bifurcations are permutation VS combination, replacement VS no replacement, and unique VS non-unique identifiability of duplicates (i.e., do balls of the same color have a serial number or are they interchangeable). I'm going to add total permutation/combination VS partial permutation/combination to the list for now, though the former may turn out to be nothing more than the special case of the latter where $r = n$. Here are the formulas I have collected:
Total Permutations
Serial Numbers, Replacement: $n^n$
Serial Numbers, No Replacement: $n!$
No Serial Numbers, Replacement: $(n - m)^n$
No Serial Numbers, No Replacement: ${n \choose u_1, u_2, u_3, ...} = \frac{n!}{u_1! u_2! u_3! ...}$
Total Combinations
Serial Numbers, Replacement: ${2n - 1 \choose n} = \frac{(2n - 1)!}{n!(n - 1)!}$
Serial Numbers, No Replacement: $1$
No Serial Numbers, Replacement: ${2n - m - 1 \choose n} = \frac{(2n - m - 1)!}{n!(n - m - 1)!}$
No Serial Numbers, No Replacement: $1$
Partial Permutations
Serial Numbers, Replacement: $n^r$
Serial Numbers, No Replacement: $\frac{n!}{(n - r)!}$
No Serial Numbers, Replacement: $(n - m)^r$
No Serial Numbers, No Replacement: ____
Partial Combinations
Serial Numbers, Replacement: ${n + r - 1 \choose r} = \frac{(n + r - 1)!}{r!(n - 1)!}$
Serial Numbers, No Replacement: ${n \choose r} = \frac{n!}{r!(n - r)!}$
No Serial Numbers, Replacement: ${n - m + r - 1 \choose r} = \frac{(n - m + r - 1)!}{r!(n - m - 1)!}$
No Serial Numbers, No Replacement: ____
The hope is that the first missing formula becomes $\frac{n!}{u_1! u_2! u_3! u_4! ...}$ and the second becomes $1$ when we plug $n$ into $r$, therefore making the total permutation/combination formulas redundant. Is everything I have thus far correct, and what formulas go in the two blanks?
Define $c=n-m$, the number of colors.
For the second blank, you could use a generating function. The desired number is the coefficient of $x^r$ in the expansion of $$ \left(1+x+\ldots+x^{u_1}\right)\left(1+x+\ldots+x^{u^2}\right)\ldots\left(1+x+\ldots+x^{u_c}\right). $$ Since $u_1+u_2+\ldots+u_c=n$, you can see that $x^n$ occurs with coefficient $1$ in this product, as you predicted.
To compute the coefficient of $x^r$ for general $r$, one can use $1+x+\ldots+x^k=\left(1-x^{k+1}\right)/(1-x)$ to write the product as $$ \left(1-x^{u_1+1}\right)\left(1-x^{u_2+1}\right)\ldots\left(1-x^{u_c+1}\right)(1-x)^{-c}. $$ By the generalized binomial theorem, $$ \begin{aligned} (1-x)^{-c}&=\sum_{r=0}^\infty(-1)^r\frac{(-c)(-c-1)\ldots(-c-r+1)}{r!}x^r\\ &=\sum_{r=0}^\infty\binom{c+r-1}{c-1}x^r. \end{aligned} $$ The coefficient of $x^r$ in this expression is the number of multisets of size $r$ with elements chosen from a set of $c$ elements and no restriction on multiplicities. Equivalently, it's the number of unordered collections of colors of size $r$, with $c$ choices of color and no limitation on the number of times a color may be used. Obviously, however, we do need to restrict the number of times a color may be used: color $1$ may be used at most $u_1$ times, color $2$ may be used at most $u_2$ times, and so on. This restriction is imposed by the other factors in the generating function: multiplying $(1-x)^{-c}$ by $\left(1-x^{u_1+1}\right)$ eliminates those multisets in which color $1$ occurs with multiplicity $u_1+1$ or higher, multiplying by $\left(1-x^{u_2+1}\right)$ eliminates those multisets in which color $2$ occurs with multiplicity $u_2+1$ or higher, and so on. It might seem that there's some double-subtraction going on here: for example, multisets in which both color $1$ and color $2$ occur with too high a multiplicity get subtracted once due to the multiplication by $\left(1-x^{u_1+1}\right)$ and get subtracted a second time due to the multiplication by $\left(1-x^{u_2+1}\right)$. The correction for this, however, is a built-in feature of the method: in the expansion of $\left(1-x^{u_1+1}\right)\left(1-x^{u_2+1}\right)$ the term $x^{u_1+1}x^{u_2+1}$ results in the adding back of the double-subtracted multisets. In fact, the entire inclusion-exclusion procedure unfolds when $\left(1-x^{u_1+1}\right)\left(1-x^{u_2+1}\right)\ldots\left(1-x^{u_c+1}\right)$ is expanded. The result is that the coefficient of $x^r$ is $$ \sum_{S\subseteq\{1,2,\ldots,c\}\atop\sum_{j\in S}(u_j+1)\le r}(-1)^{\lvert S\rvert}\binom{c-1+r-\sum_{j\in S}(u_j+1)}{c-1}. $$
One could, of course, reach this conclusion without the scaffolding of generating functions, as the only ingredients in the derivation are the formula for the number of multisets with unconstrained multiplicities and and the principle of inclusion-exclusion.
A generating function can also be used for the first blank. In this case, you need an exponential generating function. The desired number is the coefficient of $\frac{x^r}{r!}$ in the expansion of $$ \begin{aligned} &\left(1+x+\frac{x^2}{2!}+\ldots+\frac{x^{u_1}}{u_1!}\right)\left(1+x+\frac{x^2}{2!}+\ldots+\frac{x^{u_2}}{u_2!}\right)\\ &\quad\ldots\left(1+x+\frac{x^2}{2!}+\ldots+\frac{x^{u_c}}{u_c!}\right). \end{aligned} $$ The coefficient of $\frac{x^n}{n!}$ is $\frac{n!}{u_1!\,u_2!\,\ldots u_c!}$, as you predicted. I don't know of a simple expression for the coefficient of $\frac{x^r}{r!}$ in general. One thing that can be said is that if $u_j\ge r$ for all $1\le j\le c$, this case is the same as the case of partial permutations, no serial numbers, replacement, in which case the number is $c^r$. (This is because you have sufficiently many balls of every color to fill all the slots, so it doesn't matter that there's no replacement.) It's not immediately apparent from the generating function that the coefficient of $\frac{x^r}{r!}$ is $c^r$, but you can see this as follows: since $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots$, the expansion of the generating function agrees with that of $(e^x)^c=e^{cx}$ at least as far as the $x^r$ term. But the $x^r$ term in $e^{cx}$ is $c^r\frac{x^r}{r!}$, and hence the coefficient is $c^r$.