Combinatorial Problem: Groups of 5 Social Diners I am trying to figure out a way to seat 40 and 50 (n=40 or n=50) people at tables of 5 and have them rotate some number of times so that everyone meets each other.  
What is the minimum number of rotations? And what are the outcomes of combination (seating at each table for each rotation)?  What happens to rotations if missing 1 or 2 people, so n=39, 38 or n=49, 48 etc.?  
Not even sure where to begin and how to calculate rotations if combinational, or rotational if closely related to social diner or golfer problems.  Thanks!
 A: The problem of creating groups of five people out of $n$ so that every pair occurs in at least one group is called a covering problem in block designs.  This question has an additional difficulty in splitting the set of people into disjoint blocks that will be seated at the same time.  These are called resolvable block designs.
Clearly with enough "rotations" we can have every pair meet, but usually it is not possible to be so efficient that each pair meets once and only once.  In particular it is necessary that the number of people $n \equiv 5 \pmod{20}$, and it has been conjectured to be a sufficient condition as well.  While $n=45$ meets this necessary condition, existence of a resolvable block design of this kind, RBIBD(45,5,1), seems to be an open problem (Thm. 3.3, Abel, 2007), one of only four possible exceptions ($n=45,345,445,645$).
Finding the best possible/minimum "number of rotations" is generally a difficult combinatorial search exercise.  If we set aside the issue of resolvability (partitioning the blocks into $\lceil (n-1)/4 \rceil$ "parallel classes" of disjoint subsets), a good resource would be Dan Gordon's La Jolla Covering Repository tables for $t=2$ (all possible pairings), which tell us that for $n=40$, a covering of all pairs with 82 blocks of five is possible (though it is not indicated this known to be the minimum number of blocks), and that for $n=50$, a covering of 130 blocks (credited to Jan de Heer and Steve Muir) is known to be the optimal (minimum) number of blocks.
There is no indication that either of these blocks designs are resolvable, and indeed since any "course" of a dinner or "rotation" would require exactly $n/5$ blocks, it is evident the $82$ blocks for $40$ people is not resolvable (since $82$ is not divisible by $8$).  
Update (04/24/20): While $130$ is divisible by $10 = 50/5$, one can rule out the possibility of a resolvable $(50,5,1)$ BIBD with that many blocks because of the second necessary condition, that $n-1=49$ be divisible by $k-1=4$.  One needs $n \equiv 5 \bmod 20$, and $n=50$ doesn't satisfy that necessary condition (and no programming was required!).  The existence of a $(45,5,1)$ RBIBD remains an open problem (although $(45,5,2)$ RBIBDs are known to exist, where every pair of people would meet exactly twice).
