Tetrahedrons in a cube

Question

How many tetrahedrons (not necessarily regular) can be formed using the vertices of a cube?

I tried the face by face strategy, that is, I selected 3 points from each face for the base and one from the four above it. Then I multiplied it by the number of faces to get the answer as 96. Then I realised I'd missed out on the triangular bases not formed by multiple faces of the cube.

So I simply selected 4 points from the 8 vertices, giving 70. Then I subtracted all the 6 faces and the diagonal cross sections to get 58. Since the first approach gives more answers, I wanted to know why it's wrong.. Please help me with this.

Answer 1

When you did your first approach, you generated tetrahedera where three faces were coplanar with the faces of the cube. You counted those once for each coplanar face, thus triple counted them. And, as you pointed out, missed some other tetrahedra.

Correct methods of counting are given in both other answers and your own question. So I stop here.

Answer 2

By an alternative counting I obtain 58, indeed consider two opposite faces of the cube, then we can obtain all the tetrahedrons in three ways

• $3$ vertices from one face and $1$ from the opposite: $2\cdot 4\cdot 4=32$
• $2$ vertices from each faces avoiding the degenerate cases: $\binom{4}{2}\binom{4}{2}-8-2=26$

Answer 3

Here's another way: Any set of four points are vertices of a tetrahedron unless they are in the same plane. There are $8$ choose $4$ four-vertex subsets, for 70 such subsets. If no two of the four vertices share an edge, they will not be coplanar. If two of them do share an edge, they can be coplanar by being all on the same face or by being on opposite edges. There are six faces and there are six pairs of opposite edges, so after subtracting those 12 cases we are left with 58 tetrahedra, agreeing with the previous answer.