## Length of a conference

Professor Baldy has found out that 5 of his friends are attending the same conference as he. During the lectures at the conference, of these 5 people he encounters:
– every individual $$10\times$$,
– every pair $$5\times$$,
– every triple $$3\times$$,
– every quadruple $$2\times$$,
– all five $$1\times$$.

How many lectures were at the conference, given that the professor encountered at least one of his friends at each lecture?

• #### Solution

The entire group of five must have been present at one lecture. At that lecture, the professor of course encountered every group of 4, 3, 2 and 1.

We must account for the remaining quadruples, for which we must have 5 lectures, during which each triple is encountered $$2\times$$, each pair $$3\times$$ and each individual $$4\times$$.

The triples are now all accounted for.

To account for the remaining pairs we need 10 lectures, in which each individual is encountered $$4\times$$ more.

To account for the remaining individuals we need 5 more lectures.