I want to derive a formula for $S(m,m-1)$ where $S(m,n)$ is the number of ways to seat $m$ people at $n$ circular tables with at least one person at each table, The arrangements at any one table are not distinguished if one can be rotated into another and the ordering of the tables is insignificant.
So the way I think about it is breaking this into cases.
When $m$ people are seated on $m-1$ tables so that each of the tables contain at least $1$ person. There can be
Case 1
Two people are seated on one table and the remaining $m-2$ people are seated on $m-2$ tables, so that each of these $m-2$ tables have one person each. Now there are ${m \choose 2}$ ways to choose these two people but then how many ways for this case in total ? is it just ${m \choose 2}$ because we don't really care about the ordering of the tables. We just care about the two people who are seating in that table right ?
But then I can't really think of any other cases. Is it true that I only have one case here ?
So If would to make a guess, I would just say $$m \choose 2$$
Community wiki answer so the question can be marked as answered:
As Brian commented, both the result and the reasoning are correct.