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Note: not a duplicate of this question as I'd like to know what's wrong with my approach.

My attempt: Choose any 3 of the $n$ vertices of the polygon, and let the number of points between the three be $x$,$y$ and $z$. Then, $x+y+z=n-3$ and $x, y, z\ge 1$. The number of possible solutions is a sticks-and-stars problem, and can be shown to be $n-4 \choose 2$. From the top answer to the question linked above though, the correct answer is $\binom{n}{3}-(n-4)n-n$.

Putting in some values for $n$, I can see that I'm undercounting. How?

Ray Bradbury
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2 Answers2

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Your approach is nearly correct, you just need to tweak it because you are only determining the spacing between the vertices, but not the vertices concretely.

For instance, if $n=10$, then the solution $3 + 2 + 3$ means you picked 3 vertices spaced out by $3,2$ and $3$, but concretely we don't know which they are. (You basically need to pick one of the vertices, because then the others will be fixed.)

In fact, if you multiply your result by $n/3$, you'll see that you get the correct result:

$$\frac n3\binom{n-4}2 = \frac{n(n-4)(n-5)}{6} = \binom n3-(n-4)n-n.$$

Luke Collins
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  • That last line might be misleading, in that (to me) it suggests that OP is on the right path and that a minor change will get them to the correct answer. – Calvin Lin Nov 18 '20 at 21:02
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    @CalvinLin This is the case, he just needs to multiply by $n$ (fix the spacing relative to one vertex), then divide by 3 since each triple ${a,b,c}$ of vertices can be obtained in three equivalent ways, depending which vertex is chosen (the solutions of the equations are then cycled around). – Luke Collins Nov 18 '20 at 21:05
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  1. How are you setting up the bijection?

IE Say $n = 10$ and we chose vertices 2, 4, 6. What does $x, y, z$ correspond to?

What you're going for is something along the lines of

  • Fix a vertex of the triangle, let it be vertex 1. There are $n$ possibilities.
  • Let $x \geq 2 $ correspond to the difference between the index of the second vertex and vertex 1.
  • Let $y \geq 2$ correspond to the difference in the index of the third and second vertex.
  • Let $z \geq 1$ correspond to the difference between $n$ and the index of the third vertex
  • We have $x+y+z = n-1$ and $(x-1) + (y-1) + z = n-3$ as positive integer solutions. Thus, there are ${n-4 \choose 2 }$ possibilities.

Hence, there are $ n \times { n - 4 \choose 2 } $ possibilities, but we double-counted 3 times, so it is $ \frac{ n (n-4)(n-5) } { 6}$.

Calvin Lin
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