The size of a conjugacy class in the symmetric group I'm trying to prove the formula for the size of a conjugacy class in $S_n$. here is my try,
Suppose the permutation whose conjugacy class size is to be found has a number of $k_i$ cycles of lengths $m_i$ where $i$ is between one and some integer $s$ in its cycle decomposition.My idea is to fill in the places of the cycles one by one in a manner that guarantees that every integer appear only once.To that end, I find all possible ways to form sets of lengths $m_i$ from the original set of $n$ integers, and the number of these ways is (taking into account that numbers in a cycle may change order and therefore change the permutation) $$ \frac{n!}{\left(n-m_i\right)!\ m_i}$$ and once the first place is filled I have to ensure that the second cycle has completely different numbers so the number of $m_i$ cycles that may be put in the second place and that have different numbers is $$\frac{\left(n-m_i\right)!}{\left(n-2m_i\right)!\ m_i}$$ and so I proceed in this manner until all the places are filled 
once I find all possible ways to fill in every place the problem is equivalent to finding the number of possible ways to choose one element from each one of a  certain number of sets (that are disjoint) and the number of ways is the product of possible choices . and skipping a lot of computation I find the final answer to be $$\frac{n!}{\prod_{i=1}^s\left(m_i^{k_i}\right)}$$ which is wrong 
Most explanations on the web say that we must divide by $k_i!$ because the order in which the cycles are present doesn't matter but I think I've taken that into account the way I did it and still how does that imply that we must divide by the product of $k_i$ factorials? this is a big skip that I wouldn't do if I didn't have a systematic way to justify doing it. However, I'm probably wrong
please be as detailed as possible in your answer because I looked for explanations on the web but they all skip many details and I don't seem to be able to justify a lot of these skips
 A: Here's an example, with $n=10$, you can generalize it.
$(1,2,3)(4,5,6)(7,8)(9,10)$
For each of the $10!$ orderings, consider the relabeling with the cycles in the same order $3,3,2,2$. For example,
$(4,2,5),(6,8,7)(10,3)(1,9)$   corresponding to the ordering $4,2,5,6,8,7,10,3,1,9$
The only question is how many of these orderings give the same permutation. There are two issues.
1) The two $3$-cycles can be ordered in $2! = 2$ ways to give the same permutation e.g. $(6,8,7),(4,2,5)(10,3)(1,9)$
The two $2$-cycles can be ordered in $2! = 2$ ways to give the same permutation e.g. $(4,2,5),(6,8,7)(1,9,)(10,3)$
2) WITHIN each $k$-cycle, there are $k$ orderings that denote the same cycle. For our example that's a factor of $3*3*2*2$ separate from re-ordering whole cycles as in (1).
The overall factor $2!*2!*3*3*2*2$ is the number of relabelings of the original permutation which leave it unchanged as a permutation -- i.e. the number of permutations that conjugate it to itself, i.e. the number of permutations that commute with it.
So there are $10!/(2!*2!*3*3*2*2)$ distinct conjugates of $(1,2,3)(4,5,6)(7,8)(9,10)$
