4This is a solution from http://www.mathlinks.ro/viewtopic.php?t=311551
There are C420 ways to choose any 4 vertices.
Now let's count number of ways to choose 4 vertices, where at least 2 are adjacent.
1) 2 vertices are adjacent and the other 2 are not.
There are 20 ways to choose 2 adjacent vertices.
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There are 16 left places for the other 2.
We can select them in C216 ways.
15 ways are to choose 2 adjacent vertices, so there are total 20(C^{2}_{16}-15)=2100 ways in this case.
2) 2 pairs of vertices are adjacent.
There are 20 ways to choose first pair of adjacent vertices.
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There are 16 left places for the other 2 adjacent vertices.
We can select them in 15 ways.
So there are total \frac{20\cdot15}{2}=150 ways in this case. (We divided by 2 because we can get the same 2 pairs of adjacent vertices in 2 different ways).
3) 3 vertices are adjacent.
There are 20 ways to choose 3 adjacent vertices.
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There are 15 left places for the 4'th vertice, so there are total 20\cdot15=300 ways in this case.
4) 4 vertices are adjacent.
20 ways in this case.
So the answer is C420-(2100+150+300+20)=2275
