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Tetrahedra Tester

Age 14 to 16
Challenge Level Yellow starYellow star
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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

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

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