Pigeonholing finite aliens on a spaceship 
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. 

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.
 A: 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 $0$?

A: 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-) 
A: An alien can know between $0$ and $n-1$ aliens. That is $n$ possibilities. If one alien knows everyone else, then there is no one who doesn't know any one. And the other way around. So possibilities $0$ and $n-1$ are mutually exclusive. So we are down to $n-1$ possibilities for each alien to know others.
So if there are $n$ aliens and $n-1$ possibilities of knowing others, with then at least $2$ aliens exist who know the same number of aliens on the spaceship.
A: suppose there are 10 aliens for example .
alien 1 knows alien 2 , how many aliens does 2 know now ?? for the hypothesis to be false alien 2 already knows alien 1 and for him not to know just 1 lets assume he knows alien 3 thus knows 2 .
same for alien 3 , he must know 3 .. continueing onward when we get to alien 10 he knows 10  aliens but since he cant count himself thats a contradiction hence 2 aliens know same number of aliens . 
A: 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.
