Meeting management A friend wants to have ten meetings of six people every day for five days with no pair of people meeting twice. Is this possible? It appears to be a question about maximal decomposition of a complete graph on 60 vertices into sets of 10 disjoint complete graphs on 6 vertices such that no edge is used twice. Does anyone know a useful theorem or, better, a way of constructing an example?  
 A: Here is an alternate formulation which you may find easy to implement as a random search:  Find 60 five digit numbers (leadng zeroes allowed and required) such that every digit occurs equally often in each position (ones, tens, hundreds, etc.) among the collection, and any two numbers differ in their digits in at least four positions.  When you have it coded, you can try seeding it with some small sets like 00000 11111 ... 99999 01234 and see how far you get with random selection.  If you succeed in picking 20 such numbers, you may find the candidates for the 21st number to be quite limited, and in some cases show that you can't complete the list with those 20 numbers.
Gerhard "Ask Me About System Design" Paseman, 2012.02.24
A: Thanks Gerhard, that is roughly what I am trying. I get very close with four days but have not succeeded with 5. However, the numbers of permutations are very large - other wise I would go through them systematically. I thus wondered if there was some theory or a fast algorithm I could use.
Just as an interesting fact, I note that cycling round through the vertices in multiple of prime numbers (that are not factors of 60) counting off 6 for each group produces quite promising results. However, I guess there is something about working out whether numbers are mutually prime in modulo 60 that I should need to understand. 
