In problem 6 from Olimpiada Matemática Española 1996 there are 16 spies. Each spy spies some of their colleagues; moreover, given any two spies A and B, if A spies B then B does not spy A. We also know that if we take any subset of 10 spies it is possible to sort them in a chain where the first spies the second, the second the third, and so on with the tenth which spies the first. The problem asks to show that for each group of 11 spies such a chain can be made.
The solution starts with defining three numbers for each spy $A_i$. $a_i$ is the number of spies spied by $A_i$; $b_i$ is the number of spies who spy $A_i$; $c_i$ is the number of spies never spying not spied by $A_i$. Then it says that $a_i+b_i+c_i = 15$, and that's clear. But is also says that $a_i+c_i \le 8$ and $b_i+c_i\le 8$, "because otherwise it is not possible to order 10 spies as required" (if my understanding of Spanish is correct). I cannot understand the reason of this. Can somebody help me?