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

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