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:**

There are $N$ aliens, and assume one of the alien knows $K$ aliens. $(N>K)$

Can each alien know **distinct** number of aliens? 

Assume $A\_1$ knows ${A\_2,A\_3 \dots A\_{K+1}}$, now that $A\_2$ knows $K$ aliens, the aliens who he knows also knows him.(Though confusing, its the fact -evil laugh-)
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.