Number of Tetrahedrons in a Cube I am trying to find out Number of Tetrahedrons in a unit cube
My Try: Total ways to choose four vertices if $\binom{8}{4}=70$
From this we have to remove 
$1.$ All $6$ faces
$2.$ $6$ Quadrilaterals formed by face Diagonals
$3.$ $2$ Concave Quadrilaterals formed by Body Diagonals
Hence Total tetrahedrons is $70-6-6-2=56$
Is this fine?
 A: There are ${8\choose4}=70$ ways to choose $4$ vertices of the cube. Now we have to eliminate the coplanar quadruples. How many of them  are there?
We can see the $6$ quadruples coming from the faces of the cube and $6$ more coming from pairs of opposite edges of the cube. If these are all then there remain $58$ admissible quadruples.
Drawing a figure we cannot detect more coplanar quadruples. For a proof that there are indeed none we can argue as follows: The $8$ vertices of the cube can be partitioned into a "blue" and a "red" quadruple, both of them  forming a regular tetrahedron. A forbidden quadruple  would have to contain vertices of both colors, among them at least two red ones, say. These two are lying diagonally opposite on a common face of the cube. Every blue vertex of the cube is connected by an edge to one of these two red vertices; hence the plane of our forbidden quadruple contains at least one edge of the cube. It is then easy to see that this quadruple is one of the forbidden quadruples counted above.
