Final answer:
The friends and enemies problem in graph theory can be solved using the pigeonhole principle; demonstrating that within a group of 10 people, there will always be at least three mutual friends or four mutual enemies.
Step-by-step explanation:
The question you're asking relates to a classic problem in graph theory, a branch of discrete mathematics. It is known as the friends and enemies problem or a special case of Ramsey's Theorem. To solve the problem, we make use of the pigeonhole principle.
Consider a single person within this group of 10 people. This person has 9 connections with the others, and each connection is either a friendship or enmity. Apply the pigeonhole principle: if we divide the 9 connections into friends and enemies, there must be at least 5 that are friends or at least 5 that are enemies, because 9 divided by 2 is 4.5, and you can't have a half connection.
Let's assume there are 5 friends (if there are 5 enemies, the argument is analogous). Now look at these 5 friends. Each of these 5 friends has at least 4 connections with each other. Using the pigeonhole principle again, there will be at least 3 that are mutual friends or at least 3 mutual enemies within this group of 5. If there are 3 mutual friends, our condition is met with three mutual friends.
If there are not 3 mutual friends, then each of the 5 must have at least 3 enemies within this group. This means that among the 5 friends of our original person, there must be one who is an enemy to at least 3 others, creating a set of 4 mutual enemies. So, in any case, there will always be a group of three mutual friends or four mutual enemies within the group of 10 people.