Every parallelepiped has eight vertices. A tetrahedron is defined as four non-coplanar points in 3-space. How many distinct tetrahedrons can we find among the eight vertices of a parallelepiped?

I believe I know the answer, but without an answer key I don't have much confidence, and I feel a bit queasy about my reasoning. I think the answer is 60, and here's why.

First, I calculated the number of combinations of 8 things taken 4 at a time (using the binomial coefficient) and it came out 70. But, not all those combinations will be tetrahedrons because some of them contain four coplanar points and I'm using a definition that excludes these "degenerate" tetrahedra. How many should I subtract? As I visualie a parallelepiped, I picture ten planes that would contain four vertices, including six planes that contain faces and four more that contain long diagonals. I figure each one corresponds to a combination of four vertices, so I subtract ten from 70 and come up with 60.

I'm troubled that this isn't a purely combinatoric answer. Also, when I try to build up that 60 from parts, I can't work it out. I feel I must be doing something wrong.

A little background. I'm working through Norman Wildberger's WildLinAlg videos, which covers a range of topics similar to what you'd expect in a first-semester linear algebra class, though its approach is unusual. I'm doing this on my own. So far as I know, these videos are not part of any instructional program anywhere. This question is an exercise Dr. Wildberger gives at the end of Lecture 9, "3D Geometry."