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To answer the question number of paths of length 4 between vertices 1 and 2 in the complete graph $K_n$.

I think there are 3!C(n,3) paths?

because in a complete graph, all vertices are connected, that implies you can travel among any of them without any problem.

Here, a length 4 path has 5 vertices, so it implies n choose 5, however, it requires that all these paths are between vertices 1 and 2, so, 2 of the 5 points are fixed in this case.

Therefore, it is n choose 3, also, then each of the subset of these 3 vertices can permute and form different graph.

So, my conclusion is 3!C(n,3). did I miss anything?

also, do I need to multiple my answer if it is correct? since a path can go both ways.

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    However, the three interior vertices of the path must be chosen from $n-2$ candidates, not $n,$ since $1$ and $2$ are already taken. For instance, when $n=3,$ the number of paths is $0,$ not $6.$ – 2017-02-12
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    @bof thanks! so the answer is 3!C(n-2,3) ? – 2017-02-12
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    Yes, that would be my answer. As you pointed out, you'd have to double that if you want *directed* paths. – 2017-02-12

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Your logic is correct except that one would choose three vertices from among $n-2$ available vertices (you cannot revisit vertex $1$ and $2$), giving $$ 3! C(n-2,3) = (n-2)(n-3)(n-4) $$ possible paths.

You could interpret that solution in another way. Begin at vertex 1 and choose the next vertex in your path. There are $n-2$ choices available (any vertex except 1 or 2). Now choose the next vertex in your path from the remaining $n-3$ vertices, and the next from the remaining $n-4$ vertices. Now you must finish your path at vertex 2, giving you no freedom to choose. Multiplying the number of options at each step gives a total of $(n-2)(n-3)(n-4)$ paths.