**A**fter posting the question on dinner table permutations on StackExchange (mathematics), I got almost instantaneously a reply that the right number was *six*, linking to the Covering chapter by Gordan and Stinson, taken from the second edition of *Handbook of Combinatorial Designs*. If I understand correctly this terse coverage that reminds me of Gradshteyn and Ryzhik’s monumental *Table of Integrals, Series and Products*, my question is connected to the covering number *C(20,5,2)*, which is equal to *21* according to Table 1.33 of the above chapter. This is the minimum number of blocks such that every pair of points occurs in at least one block (element) of the collection of subsets of size 5 (the tables across courses). Now, in my problem, the interesting coverage is a *resolvable coverage*, i.e., a collection of tables that “can be partitioned into parallel classes, each of which consists of *v/k*[=4] disjoint blocks” (Definition 1.43). Then the number of parallel classes is *r(4,5)=6*, as provided by hardmath. Now the remaining question is how to build the resolvable 2-(20,5,1) covering, i.e. the table plans across the six courses… This may be related to a direct proof as to why a five course dinner does not work.

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