I have this given problem:
In a class there is 6 students,every 2 students of 6 know each other in advance or they don't.
Show that there is 3 students that don't know each other in advance or that there is 3 that know each other in advance.
This problem looks simple but when i try to write a solution for it i'm confused if it's enough or not, here is my attempted:
Case (every two students know each other in advance):
Let $G=(V,E)$ be a graph with $|V|=6$ vertices ,to represent this case the number of edges in this graph should be $|E|=3$. By using the pigeon hole principle we want to place $|V|=6$ pigeons(vertices) in $|E|=3$ holes(edges) where at each hole there is exactly 2 pigeons.
if we choose a pigeon(vertex) $u\in{V}$ at hole $a\in{E}$ then this ensures that there is 3 other vertices in different holes that are unreachable from $u$.
Is this solution correct or even enough? for the second case i'm planning to use the same reasoning but I plan to let the vertices be the holes ,and edges for the pigeons.
Pick one student. Among the five others, there must be three he knows, or three he doesn't know. Suppose he knows three people. If two of them know one another, we have a group of three mutual acquaintances -- those two and the original student. If none of the three knows one another we have a group of three mutual strangers.