Triangles In a Cube !!!

We have 8 vertices and we need to form a triangle with 3 vertices. Thus, there are exactly $\binom{8}{3}$ different triangles.
On a face of a cube, there are 4 vertices. So, we can form $\binom{4}{3}$ different triangles on a face. We have 6 faces, so there are $6 * \binom{4}{3}$ triangles whose vertices lie on the same face. We subtract them from the number of all different triangles and we get our result:

$\binom{8}{3} - 6 * \binom{4}{3} = 56 - 6 * 4 = \boxed{32}$

How many triangles can I forme use all the vertices? $\binom{8}{3}$

How many triangles can I forme use the vertices of one face? $\binom{4}{3}$

Total is $\binom{8}{3}-6\times\binom{4}{3}=\boxed{32}$

Thats 56 - 6x4 = 32