Combinatorics Problem on a 120 sided die The Dice Lab has used the disdyakis triacontahedron to mass market an injection moulded 120 sided die. It has 62 vertices, 180 edges and 120 faces that are identical scalene triangles. A friend gave me one as a gift. The vertices are distributed along 15 great circles with 12 on each one. (15*12= 180 edges). It appears to have 3 kinds of vertex. Some (12) are intersections of 5 great circles. Some (20) are intersections of 3 great circles. And finally, some (30) are intersections of just 2 great circles. (12+20+30= 62 vertices).
There are 62 "choose" 3 written mathematically $\binom {62}{3}$ = 37820 ways to pick 3 vertices from those available. 120 of those ways are the vertices of the face triangles. 
My question is this: How many of those ways have all 3 vertices on distinct great circles?
Combinatorics is my worst topic in mathematics.
 A: The condition on the placement of the three vertices is:

No pair among the three vertices may share one of the 15 great circles of the 120-sided dice.

We are going to partition the vertex selections according to the degree of the selected vertices:


*

*A degree-4 vertex (where 2 great circles meet) will be called a rhomb. There are 30 of them.

*A degree-6 vertex (3 circles) is a delt (20).

*A degree-10 vertex (5 circles) is a pent (12).


I call the image below the astrolabe. It shows a hemisphere of the disdyakis triacontahedron, with vertices labelled as follows: hollow circle for a rhomb, solid circle for a pent and no circle for a delt. The other hemisphere looks exactly the same and the border shows how the vertices of one great circle are arranged. There are many other properties which I won't go into – suffice to say that it should help in visualising the extremely non-trivial calculations I present below.

Now we get into the actual computations. When we select a vertex, we kill (prevent from selecting) all other vertices on all the great circles it is part of. After such a selection, the vertices not killed are termed survivors.
Selections with a pent. If a pent is selected, we end up killing all vertices save for a ring of 10 rhombs. (Think of the selected vertex as a pole and the great circles as lines of longitude; the ring of survivors is the equator.) These 10 rhombs split into five disjoint and identical groups of two, each of which is part of an octahedron; within an octahedron, selecting one vertex kills all the others. (The entire disdyakis triacontahedron, as a spherical tiling, is a compound of five octahedra.) Hence if we are to select a pent, we must select two different groups and then one rhomb in each group:
$$\underbrace{12}_{\text{pents}}\cdot\underbrace{\binom52}_{\text{group choice}}\cdot\underbrace{2\cdot2}_{\text{vertex choice}}=480$$
Selections with two rhombs and one delt. Suppose the first selection is a delt. Then the 18 surviving rhombs can be grouped into two octahedra and three antipodal pairs, all of which are disjoint. Again, making a valid selection here after choosing the delt boils down to:


*

*choosing two groups

*selecting one vertex in each of the chosen groups


The number of ways to do this is
$$\underbrace{20}_{\text{delts}}(
\underbrace{1\cdot6^2}_{\text{oct, oct}}+
\underbrace{6\cdot6\cdot2}_{\text{oct, pair}}+
\underbrace{3\cdot2^2}_{\text{pair, pair}})=2400$$
Selections with three rhombs. As mentioned before, the disdyakis triacontehedron can be viewed as five disjoint octahedra, the 30 rhombs split into these groups of six and selecting one rhomb kills all the others in its group. To make a selection with three rhombs, we select three groups and then one vertex in each group, and the number of ways for this is
$$\underbrace{\binom53}_{\text{group choice}}\cdot\underbrace{6^3}_{\text{vertex choice}}=2160$$
Selections with two delts and one rhomb. After selecting one rhomb, there are 12 surviving delts, four of which lie on the equator (taking the selected rhomb as a pole – see above) and eight of which do not. Within both these groups, all delts are equivalent. If an equatorial delt is selected, there remain four surviving delts we can select; if a non-equatorial one is selected, there remain eight. Then we have to divide the sum of ways by two because we counted the permutations of each combination of two admissible delts.
$$\underbrace{30}_{\text{rhombs}}(\underbrace{4\cdot4}_{\text{equator delt}}+\underbrace{8\cdot8}_{\text{non-equator delt}})\cdot\underbrace{\frac12}_{\text{perms}}=1200$$
Selections with three delts. The 20 delts are the vertices of a dodecahedron. After selecting one delt, 12 delts survive, all of which are equivalent. After selecting a second delt, no matter how the first and second delts are positioned, there always remain six surviving delts that can be selected. As with the previous case, we have to divide by six because we counted permutations:
$$\underbrace{(20\cdot12\cdot6)}_{\text{delt selection}}\cdot\underbrace{\frac16}_{\text{perms}}=240$$
Conclusion. We have exhaustively enumerated above the number of valid selections by partitioning them based on the degrees of the selected vertices. It remains to sum:
$$480+2400+2160+1200+240=\mathbf{6480}$$
