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I guess the answer is: $C(n,n-3)/2$ since for $n$-sided polygon,there are $n$ vertices, and for each vertex, it cannot form diagonal with the adjacent points and itself. So, each vertex can form $n-3$ points. And half of the vertices have repeated diagonals and we need to divide by $2$.

It holds true for quadrilateral and pentagon,but why it does not fit triangles?

Michael Hardy
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JimfyWinsy
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  • Note that "vertex" is the singular, e.g. "This vertex is..." or "These vertices are...". It's like "This matrix is...." and "These matrices are....". The same applies to "vortex/vortices" and "helix/helices". I corrected it in the posting. But you should get $n(n-3)/2$, not $C(n,n-3)/2$. ${}\qquad{}$ – Michael Hardy May 07 '15 at 15:12

2 Answers2

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The number you have computed, $C(n,n-3)/2$, which I am interpreting as $\binom{n}{n-3}/2$ though this doesn't coincide with your argument, is too large. A different way:

We can choose any two points, except adjacent points are not allowed. There are $\binom{n}{2}$ pairs of points and $n$ pairs of adjacent points, so the answer is $\binom{n}{2}-n$.

Matt Samuel
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  • For clarification to those people reading and comparing answers, $\binom{n}{2} - n = \frac{n^2-n}{2} - \frac{2n}{2} = \frac{n^2-3n}{2} = \frac{n(n-3)}{2}$ and so both answers given agree. – JMoravitz May 07 '15 at 15:09
  • But what's wrong with my reasoning?We just use different ways to think about the question. – JimfyWinsy May 08 '15 at 04:50
  • @JimfyWindsy nothing's wrong with it as I said in my comment below. Just the final answer was incorrect. – Matt Samuel May 08 '15 at 13:45
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Your reasoning, which is correct, implies that the number of diagonals is $$\frac{n(n-3)}2$$ and this fits for triangles, that have no diagonals.

ajotatxe
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