Enumerating un-labeled Hasse diagrams. As a set of homework problems I've been tasked with figuring out haw many posets can be listed that satisfy some given condition.
For example: how many Join semilattices are there with 5 elements?
Going with this example, I've been going through all the possible "forms" of the Hasse diagrams that fit the condition.
Now, I need to figure out how many ways there are to label each diagram.
Take the following Hasse diagram:
    .
   / \
  .   .
  |   |
  .   .

Does anyone have a general strategy that I can use to enumerate the labelings. I'm not necessarily looking for a closed form formula or a recurrence or anything, just some tips on how to think about it. Although having a recurrence/formula would obviously be useful in providing insight.
Edit:
The comments state that the answer can be found with: n! / the number of symmetries, but what about the following diagram?
    . 
  ....   

I didn't draw the lines, but it's the 5 element poset with 4 symmetries.
I'm pretty sure the answer is 5 here. You have 5 different choices for the max element but beyond that the ordering doesn't matter.
However, I see 4 symmetries in the graph and 5! / 4 = 30.
Can someone explain?
 A: I will answer the comment in the comments, which to me seems like a reasonable answer to the entire question:
Without appealing to group theory, we can see by elementary means that the diagram you listed above must have 60 distinct labelings.


*

*Some element must be the maximum element. There are $5$ choices.

*Some pair of remaining elements must be in the middle layer. However, since we cannot distinguish the "left" from the "right" leg, it only matters which two elements we put into this pair, not which one goes where. So there are $(4\times 3)/2=6$ choices.

*Finally, we must put the last two elements in the remaining layer. This time, though, it does matter which one we put in the right leg and which we put in the left; the labelled poset can distinguish between them based on where we put the elements in the middle layer.  There are $2\times 1=2$ choices.


All of these choices are indepenedent, so you have $5\times 6\times 2=60$ distinct labelings.
(If the last step makes you uncomfortable, your concerns may be assuaged by being a bit more explicit. Without loss of generality, suppose that your label set is $\{1,2,3,4,5\}$ and in step 2, put the smaller label in the left leg. Now you can also distinguish between right and left without appealing to the geometry.)
