Originally Posted by

**Grandad** What also became clear, is that each selection of 3 questions - call them questions

and

- involves the elimination of 3 pairs -

- from the

pairs that can be chosen from the

questions. This is because no other participant can be given any other pair of questions from

and

once this group of three has been allocated to someone.

Following up on that insight of Grandad, each triple that is selected eliminates 3 pairs of questions. Since there are pairs, the maximum possible number of participants is . Grandad's computations show that this maximum is attained when n=3 and when n=7. I believe that the maximum can be attained whenever n is of the form . (In fact, I'm sure of that.) I don't have a neat way of describing the construction, but I have written it out in detail for n=15, using these 35 triples:

Code:

1 2 3 2 4 6 3 4 7 4 8 12 6 8 14
1 4 5 2 5 7 3 5 6 4 9 13 6 9 15
1 6 7 2 8 10 3 8 11 4 10 14 6 10 12
1 8 9 2 9 11 3 9 10 4 11 15 6 11 13
1 10 11 2 12 14 3 12 15 5 8 13 7 8 15
1 12 13 2 13 15 3 13 14 5 9 14 7 9 12
1 14 15 5 10 15 7 10 13
5 11 12 7 11 14

I suspect that the maximum is not going to be attained except at numbers of the form , and I would guess that there will not be a simple formula for the answer at a general value of n.