We received the following solution from Ben:
Using the combination nCr,
6C3 = all the possible triangles = 20.
If you take each possible triangle as a "base", then use nPr to find the number of tetrahedra that can be constructed from one base.
3P3 = 6
We must realize that using these figures, one tetrahedron will be listed four times, each time with a different triangle on the base, but still the same tetrahedron. Therefore, we must divide the figure by 4.
The lengths 4, 5 and 9 cannot form a triangle, so tetrahedra with this value must be removed.
There are 6 tetrahedra with base 4, 5 and 9, so there 4 times (24) this value listed. Therefore, we must subtract this value from the figure.
6 x 20 = 120 = all hypothetical tetrahedra
120 - 24 = 96 = all possible tetrahedra (inc. tetrahedra counted more than once)
96 / 4 = 24 = total number of tetrahedra
that can be made
ans = 24