I want to preface this by saying that I'm learning about combinatorics for the first time and so my terminology may be problematic.
I am trying to figure out how to count the permutations of a set given that we identify certain arrangements. Here is how I am trying to formalize this:
Let $G = \{a_{1},...,a_{n}\}$ be a set, and let $f: G \rightarrow G$ and quotient the set by this action, meaning identify $a_{i}$ with $f(a_{i})$, and let $G/f$ denote the resulting set. Suppose this action has order $m$. Then I believe the permutations of $G/f = \frac{P(n,r)}{m!}$
(Note: $P(n,r)$ denotes the number of $r$ elements permutations possible from a set of $n$ elements.)
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Here is the motivation:
Suppose we had an even number of elements in out set, and our action was reflection, meaning we identified every point to its antipodal point. Then the number of permutations must be divided by two, because we double count each permutation exactly twice.
Similarly, if we had some action that resulted in $3$ points in the same orbit, then we would have to divide by $3$, because the permutations $a_{1},...,a_{2},...a_{3} = a_{1},...,a_{3},...,a_{2}$ and etc.
...
Is my intuition correct, and if not, where is my error?
The number of permutations of $\{a_1,...,a_n\}$ upto the relation is: $$\frac{n!}{|f^{-1}(a_1)| ! \times |f^{-1}(a_2)| ! ... \times ... |f^{-1}(a_n)| !}.$$
This is because we identify all elements corresponding to $f^{-1}(a_i)$ as single element.
Let $|f^{-1}(a_i)| = m$ for all $1 \leq i \leq n$. What if we want $r$ elements out of $n$ elements: The number of permutations of $r$ elements out of $n$ elements from $\{a_1,...,a_n\}$ with the identification of relation: $$\sum_{\ell = 1}^n \sum_{\sum_{k=1}^{\ell} j_k = r, 1 \leq j_k \leq m}{|Image(f)| \choose \ell} \times\ell! \times \prod_{k = 1}^{\ell} {r - \sum_{i = 1}^{k-1} j_i \choose j_k}$$
This is because there are ${|Image(f)| \choose \ell}$ ways to pick $\ell$ equivalent sets and $\ell!$ permutations among these $\ell$ sets. From these $\ell$ sets we pick $j_1,j_2,...j_{\ell}$ elements each and there are $\prod_{k = 1}^{\ell} {r - \sum_{i = 1}^{k-1} j_i \choose j_k}$ ways of placing these equivalent elements picked from $\ell$ sets in $r$ locations.
$$ = \sum_{\ell = 1}^n \sum_{\sum_{k=1}^{\ell} j_k = r, 1 \leq j_k \leq m}P(|Image(f)|,\ell) \times \prod_{k = 1}^{\ell} {r - \sum_{i = 1}^{k-1} j_i \choose j_k}$$