Dear John, you wrote

> 2) Then we need to count the number of ways to chop our n-element set into k subsets of size p and n−kp subsets of size 1. The answer is the multinomial coefficient (np,p,…,p,1,1,…,1) with k p's and n−kp 1's.

Since the arrangement of the orbits of size 1 does not matter, I would suggest that we should count the number of ways to chop our n-element set into k subsets of size p and 1 subset of size n-kp instead. For example, consider the \$$\mathbb{Z}_2\$$-action of a set of \$$n=4\$$ elements. The number of different ways to chop the set into \$$k=1\$$ subset of size \$$p=2\$$ and \$$(n-k)=2\$$ subsets of size 1 is \$${4 \choose 2, 1, 1}=\frac{4!}{2!}=12 \$$

$$\begin{array}{cc} (12)(3)(4)&(12)(4)(3)\\\\ (13)(2)(4)&(13)(4)(2)\\\\ (14)(2)(3)&(14)(3)(2)\\\\ (23)(1)(4)&(23)(4)(1)\\\\ (24)(1)(3)&(24)(3)(1)\\\\ (34)(1)(2)&(34)(2)(1)\\\\ \end{array}$$

However, the 12 different partitions above only produce 6 different permutations, which is the same as \$${4 \choose 2, 2}=\frac{4!}{2!2!}\$$, the number of ways to chop the 4-element set into a subset of size 2 (the 2 elements that we swap) and another subset of size 2 (the other 2 elements that stay fixed). So in general, I believe the number of ways in step 2 of [your comment](https://forum.azimuthproject.org/discussion/comment/19309/#Comment_19309) should be \$${n \choose p, ..., p, (n-kp)}=\frac{n!}{(p!)^k (n-kp)!} \$$.

Finally, in step 3 of [your comment](https://forum.azimuthproject.org/discussion/comment/19309/#Comment_19309), I think not only we should include a factor of \$$(p-1)!\$$ for each orbit of size \$$p\$$ but also we should divide it by \$$k!\$$ for the fact that the arrangement of the \$$k\$$ orbits of size \$$p\$$ does not matter. For example, consider chopping the \$$n=4\$$-element into \$$k=2\$$ subsets of size \$$p=2\$$, we have

$$\begin{array}{c} (12)(34)\\\\ (13)(24)\\\\ (14)(23)\\\\ (23)(14)\\\\ (24)(13)\\\\ (34)(12)\\\\ \end{array}$$

However, the pairs \$$\left\lbrace(12)(34) \text{ and } (34)(12)\right\rbrace\$$, \$$\left\lbrace(13)(24) \text{ and } (24)(13)\right\rbrace\$$, and \$$\left\lbrace(14)(23) \text{ and } (23)(14)\right\rbrace\$$ each produce the same permutation, so the actual number of different permutations is \$$3=(2-1)!\frac{1}{2!}{4 \choose 2,2}\$$. Therefore, in general, I believe that each summand in the final formula should be

$$(p-1)!\frac{1}{k!}{n \choose p, ..., p,(n-kp)}=\frac{n!}{k!p^k(n-kp)!}.$$