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suppose that two people are either both friends with each other or both enemies of each other. in any group of 10 people, show that there are always either three mutual friends or four mutual enemies.

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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.

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