How could you arrange 36 trees into 9 rows of 8? (Note that a row is a straight line that can go in any direction.)

  • $\begingroup$ Possible duplicate of 7 Trees, 6 Rows, 3 Per Row? $\endgroup$ – dcfyj Feb 27 '17 at 17:23
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    $\begingroup$ This doesn't seem like a duplicate, just a similar class of problem $\endgroup$ – Sconibulus Feb 27 '17 at 17:24
  • $\begingroup$ Welcome to Puzzling! I've edited your question to make it clearer what you mean - feel free to edit back if I've accidentally changed it away from your intended question! $\endgroup$ – Deusovi Feb 27 '17 at 17:33
  • $\begingroup$ Thanks Deusovi. Thats exactly what I mean in my question. And its not a duplicate dcfyj i tried. I hope i get an answer for that math problem. $\endgroup$ – Aaron Feb 27 '17 at 17:40

Draw nine lines so that no two are parallel and no three meet in the same place, then place a tree at each intersection point. Every pair of lines intersects, so there are $\binom{9}{2}=36$ trees, each row being one of the nine lines.


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