Confusion about size of conjugacy class formula in $S_n$

Question

I was reading this answer The size of a conjugacy class in the symmetric group but I didn't understand point 1). Why does this contribute a factor of $2!\times2!$? After all, the 4 cycles can be moved around freely: the cycle type $(3,3,2,2)$ is an unordered list. Shouldn't the factor be $4!$?

Answer 1

Conjugate permutations have the same shape, and permutations of the same shape are conjugate. This is a trivial consequence of the fact that if $\sigma$ sends letter $a$ to letter $b$ then $\tau\sigma\tau^{-1}$ sends letter $\tau(a)$ to letter $\tau(b)$.

Hence the size of a conjugacy class is the number of permutations of that shape.

For example: How many different permutations $(abc)(def)(gh)(ij)$ are there?

Well there are $10!$ ways to list $a, \dots, j$ in order; but this overcounts.

(i) We can write $(abc)=(bca)=(cab)$ and so we must divide by $3$. The same is true for $(def)$, and also similarly for $(gh)$ and $(ij)$; so in fact we need to divide by $3^2 \cdot 2^2$.

(ii) Also we always have $(abc)(def)=(def)(abc)$ (these are disjoint cycles of course), so we also need to divide by $2!$. The same is true for $(gh)(ij)$ so again we need to divide by $2!$.

Putting it all together there are $\dfrac{10!}{3^2\cdot 2!\cdot 2^2\cdot 2!}$ such permutations.