# Number of Regular tetrahedron from Unit cube

Can we count number of Regular tetrahedrons formed out of Unit cube?

If vertices of Unit cube are taken as $(0,0,0,)$, $(0,0,1)$, $(0,1,1)$, $(0,1,0)$, $(1,0,0)$, $(1,0,1)$, $(1,1,1)$ and $(1,1,0)$

I have seen a post where any four vertices satisfying $x \le y \le z$ forms a Regular Tetrahedron.

But if i choose vertices as $(0,0,0)$, $(1,1,1)$,$(0,0,1)$,$(0, 1,1)$

will it form Regular tet?

Can i know any logical way to choose vertices of cube to form Tet?

No, because, for instance, the distance from $(0,0,0)$ to $(1,1,1)$ is different from the distance from $(0,0,0)$ to $(0,0,1)$.
You'll get a tetrahedron if you take those vertices with an odd number of $0$'s and another one if you take those vertices with an even number of $0$'s.
• Logically will total possible Tets are they $58$ Hidden in cube? Since total ways to choose $4$ vertices are $\binom{8}{4}$ and there are $6$ faces and $6$ planes formed by two Opp Edges , hence Total = $70-6-6=58$ – Ekaveera Kumar Sharma Sep 16 '18 at 16:17
Any other choice of 4 vertices from the vertex set of a cube clearly combinatorically could be structured to become a simplex as well. But those simplices would not be regular tetrahedra (as those would use at least 2 different edge length out of $\{1, \sqrt{2}, \sqrt{3}\}$, i.e. edge of cube, diagonal of square face, body diagonal of cube. In some cases such simplices even would become degenerate (when choosing all 4 vertices from a single square face).