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Exercise 6.4 · Q11

Q.Twenty points no four of which are coplanar are in space. How many triangles do they determine? How many planes? How many tetrahedrons?

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With no 3 points collinear and no 4 points coplanar, every 3 of the 20 points fix a unique triangle and a unique plane, and every 4 points fix a unique tetrahedron.

nCr=n!r!(n−r)!^nC_r=\dfrac{n!}{r!(n-r)!} — the number of ways to choose rr objects from nn, order not mattering. Here n=20n=20; triangles/planes need groups of r=3r=3 points, tetrahedrons need groups of r=4r=4 points.

  1. Triangles. Any 3 of the 20 points (no 3 collinear, so no degenerate triple) determine exactly one triangle. Number of triangles =20C3= {}^{20}C_3.
  2. Compute 20C3=20×19×183×2×1=68406=1140^{20}C_3=\dfrac{20\times19\times18}{3\times2\times1}=\dfrac{6840}{6}=1140.
  3. Planes. Any 3 points determine a plane. Because no 4 of the given points are coplanar, no plane can contain a fourth point of the set — so no two different triples can determine the same plane. Hence the number of distinct planes equals the number of triples: 20C3=1140^{20}C_3=1140. …

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