counting the number of elements in a conjugacy class of $S_n$

Question

I want to know if there is some systematic way (using some combinatorial argument) to find the number of elements of conjugacy classes of $S_n$ for some given $n$.

For example, let's consider $S_5$. If the representative for the conjugacy class is an $m$-cycle then Dummit and Foote gives a formula on how to compute the number of elements in the conjugacy class. This is not a problem. But what about when the representative is not an $m$-cycle. As an example we can consider the conjugacy class that gives rise by the partition $2+3$ of $5$. A representative for the conjugacy class would be $(1 2)(3 4 5)$. How can I find the number of such elements?.

Question?:

Does $ {5\choose 2}\cdot { 3 \choose 3}\cdot 2$ give me what I want?

Reasoning: For the first parenthesis I need to choose $2$ elements out of $5$ and for the second set of parenthesis I need to choose $3$ out of the remaining $3$ (noting that they can't be repeats). Finally we can permute these two parenthesis in two ways, thus giving me the above number.

Is this reasoning correct?. If not how does one find the number of elements of such conjugacy classes.

As always, any help is greatly appreciated.

Answer 1

More systematically, you have $n!$ choices to arrange $1,\ldots, n$. Place them into the parentheses pattern in this order to obtain an element of the conjugacy class. For each $r$-cycle, you divide by $r$ as only the cyclic order within a cycle plays a role, not which element we start with. Then, if there are $n_r$ cycles of length $r$, divide by $n_r!$ as the order in which the cycles are listed is not important. Note that this must also be done for the cycles of length $1$!

This gives us $$ \frac{n!}{\prod_{r}r^{n_r}n_r!} $$ Thus in $S_5$, there are $\frac{5!}{2\cdot 3}$ conjugates of $(1\,2)(3\,4\,5)$. Similarly, there are $\frac{7!}{2\cdot2\cdot 2!\cdot 3!}$ conjugates of $(1\,2)(3\,4)$ in $S_7$.