How many different one-to-one acquaintances does the company of $N$-persons have, if each group of three contains both acquaintances and unfamiliar people?
I have tried to calculate in this way: if the company is divided into some three subgroups, the number of them would be $N/3$. As a result, each three could have $6$ different combination ways of acquaintances. So, answer would be $6^{N/3}$. However, my teacher has said that my answer was incorrect. And it would be better to use graph theory to deal with this problem. I also know famous "Six degrees of separation" theorem. But, how to calculate for three persons?
Imagine the complete graph on $N$ vertices. Each person is a vertex. We color an edge red if it connects two acquaintances and blue if it connects unfamiliar people. The statement that every group of three has both tells us that there is no monochromatic triangle. It is a classic result that if $N\ge 6$ there will be a monochromatic triangle, so we know $N \le 5$. There is not enough information to say more.