Length of orbits in group actions Let $G$ be a group, with $|G|=40,$ and $X$ a set with $|X|=67$ and $G$ act on $X.$
There are $5$ orbits for this action of $G$ on $X.$
How many lists of orbits lengths there are?
I know that the leght of an orbit divides the order of the group $G$ and that the sum of orbits length is equal the order of the set $X$, but I don't know how to solve the problem.
 A: Here's some Mathematica code to get the possible lists of orbits:
IntegerPartitions[67, {5}, Divisors[40]]

There are six such partitions, and so there are six possible actions here.
To do this by hand, it's not that bad to just try some things out (although obviously proofs by enumeration/exhaustion are annoying).  First of all, note that $67/4=13.5$.  This means that the list can't consist of orbits of size less than 10. So, start with a list consisting of orbits no larger than 20.  Since $20+10+10+10+10 = 60 < 67$ but $20+20+10+10+10 = 70$, it's clear that there needs to be at least one orbit of size 40 or two orbits of size 20.
From there, it's a matter of trying things out.  For instance, for a list with exactly two orbits of size 20 (and none of size 40), we need the three final orbits to sum to $67-40=27$.  Quickly playing around with the numbers 1, 2, 4, 5, 8, and 10 yields nothing that works: since $8+8+8=24$, there needs to be at least one 10, so can we find two orbits that sum to 17?  Clearly not.  Etc.
