Old olympiad in Macedonia 1974.

[rsoldo]

On each side of the cube are given $n$ points, so that no three are collinear.
a) How many straight lines who don't lie on the sides of the cube are determined by the given points?
b) How many triangles who don't lie on the same side of the cube are determined by these points?
c) If we paint the sides of the cube with different colors, how many tetrahedrons with vertices at the given points we have such that:
i) three vertices of the same color;
ii) two vertices are of the same color, and the other two are of the same ( other ) color?

[rsoldo]

My results:
a) $15n^2$;
b)${n\choose 2}\cdot 5n\cdot6+n\cdot5n\cdot4n\cdot\frac{1}{2}=25n^3-15n^2$;
c)
i) $6\cdot {n\choose 3}\cdot 5n=5\cdot n^2 (n-1) (n-2)$;
ii) $\frac{6\cdot {n\choose 2}\cdot{n\choose 2}\cdot 5}{2}=\frac{15}{4}\cdot n^2\cdot (n-1)^2$.

P.S. If anyone has any correction I would be grateful.

[Dendr]

It looks like answer for (b) is incorrect. Check for $n=1$.
$\binom{6n}{3}-6\binom{n}{3}=35n^3-15n^2$
(All possible tirangles minus the triangles lying on each separate face)