Probability of a right angle triangle being inside a regular octahedron Having a regular octahedron I choose 3 vertices from it randomly. What's the probability of those 3 vertices being vertices of a right angle triangle?
Here's how I did:


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*Having 6 vertices present in an octahedron, I get to know all of the possible cases: 6C3

*Know how many faces does an octahedron has, 8.

*In each faces there are 3 vertices, so that makes 3C3

*$\frac{8  *  ^3C_3}{^6C_3}$ = $\frac{2}{5}$


However, this is wrong, because the face of the octahedron is an equilateral triangle. How do I make this right? Thanks.
 A: The faces of the octahedron are equilateral triangles, but you can take a cross-section through the center of the octahedron that contains four vertices arranged in a square. Any three of those vertices form a right triangle.
There are $3$ such cross-sections, and in each cross-section $4$ ways to choose three vertices, for a total of $12$ right triangles.
Alternatively, notice that the three vertices include two diametrically opposite vertices; there are $3$ such pairs of vertices in the entire octahedron, and for each pair there is a choice of $4$ vertices for the third vertex, and in all such cases (and no others) we have a right triangle, so there are $12$ right triangles.
Alternatively, take any of the $6$ vertices as the right angle; then there are only $2$ other pairs of diametrically opposed vertices that can be the other two vertices of a right triangle, for a total of $12$ right triangles.
Alternatively, observe that it is not possible to choose three vertices of which none are adjacent. Suppose we have chosen any three vertices;
identify two of these vertices that are adjacent,
and consider where the third vertex might be.
There are four possible locations, of which two produce faces and two produce right triangles.
Therefore any three vertices (which can be chosen in $\binom63 = 20$ ways)
are either vertices of a face or vertices of a right triangle.
There are $8$ faces, which leaves $20 - 8 = 12$ right triangles.
