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

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

Taking these into consideration, your answer and my answer will match.