While reading the explanation of the identities involving the Stirling numbers of the first kind, I encounter this problem that usually confuses me:
In the first argument that is marked in red, because we explicitly single out 3 groups of 2 elements in ${6 \choose 2,2,2}$, we have to take care of the issue of double counting them, hence the division by 6.
The problem I usually have is in the second argument marked in red: when we focus on choosing the groups of 3 elements, as in this case, why don't we have to care about their order relative to the remaining group of 2 elements? i.e. why don't we have do divide by 2 in this case?
By extension, I don't understand why, in the first argument, the orders of the $(n-6)$ fixed points and the ${6 \choose 2,2,2} \times \frac 16$ three two-cycles are not considered. And in the second argument, why the orders of $(n-5)$ fixed points and the three-cycle and two-cycle are not considered.
But we do. Note that $\binom53$ stands for $\frac{5!}{3!2!}$ so we divide by $3!$ and also by $2!$. You might think that multiplication $\dots\times2$ that follows is an effort to neutralize, but it is not. It corresponds with the number of $3$-cycles that are possible. Working with e.g. $1,2,3$ there are $2$ of such cycles: $(1,2,3)$ and $(1,3,2)$.
Further the orders of fixed points are not considered because e.g. $(1)(2)$ is exactly the same permutation as $(2)(1)$ and we only want to count it once. The order of them does not affect the permutation.