SOLUTION TO PROBLEM R-13

There are $\binom{16}{3}=560$ sets of 3 points. We must exclude from our count those sets of three points that are collinear. There are 4 vertical and 4 horizontal lines of four points each. These 8 lines contain $8\cdot\binom{4}{3}=32$ sets of 3 collinear points. Similarly, there are $2\cdot\binom{4}{3}+4\cdot\binom{3}{3}=8+4=12$ sets of 3 collinear points that determine lines of slope $\pm 1$. Since there are no other sets of 3 collinear points, the number of triangles is 560 - 32- 12 = 516.

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Apr 19 08:44:22 CDT 2002