Question: > 
> There is a group of finite aliens on a spaceship. Show that there are at least $2$ aliens who know the same number of aliens on the spaceship. 
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I was given a hint, and that was to use the pigeonhole principle. I think I can visually see it but I am unsure how to show it.

Answer: 
HINT: Let $n$ be the number of aliens on the ship. As noted in the comments, you must assume that $n\ge 2$. Suppose that A is an alien on the ship; how many of the others can it know? The largest possible answer is $n-1$, and the smallest is $0$. 

* How many possibilities is that?
* Is it actually possible for one of the aliens to know $n-1$ of the others and another to know $0$?
consider lattice points in the plane. an example is say there are 4 aliens. just like a dice these 4 dots must have two lines or else two aliens would Know O others. then suppose 1 alien knows 1 other and the other 3 must know a different number of the aliens left . this means one dot will have 2 lines , one will have 3 and in all we would have 1,2,3, lines respectively. Its clear then these lines will connect to the other dots hence a contradiction. the case of n=4 , n=5 ect can be generalized to saying N dots must have only (n-1)n/2 lines but this sum by pigeonhole means 2 aliens will thus have the same number of lines connected to them.