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How many triangles can be formed by the vertices of a regular polygon of $n$ sides? And how many if no side of the polygon is to be a side of any triangle ?

I have no idea where I should start to think. Can anyone give me some insight ?


Use Combination or Permutation

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    $\begingroup$ When all else fails, make sure you have a clear understanding of the definitions and do some small examples. $\endgroup$ – hardmath Sep 22 '15 at 12:56
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    $\begingroup$ Think about the vertices of the polygon as potential candidates for vertices of the triangle. Using that, you get (n choose 3) as the number of possible triangles that can be formed by the vertices of a regular polygon of n sides. $\endgroup$ – StopReadingThisUsername Sep 22 '15 at 13:09
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Notice,

Total number of triangles formed by joining the vertices of n-sided polygon $$N=\text{number of ways of selecting 3 vertices out of n}=\color{}{\binom{n}{3}}$$ $$N=\color{red}{\frac{n(n-1)(n-2)}{6}}$$ $\forall \ \ \color{blue}{n\geq 3}$

Number of triangles having one side common with that of the polygon $$N_1=(n-4)n$$ Number of triangles having two sides common with that of the polygon $$N_2=n$$ If $N_0$ is the number of triangles having no side common with that of the polygon then we have $$N=N_0+N_1+N_2$$ $$N_0=N-N_1-N_2$$ $$=\binom{n}{3}-(n-4)n-n$$ $$=\color{}{\frac{n(n-1)(n-2)}{6}-n^2+3n}$$ $$N_0=\color{red}{\frac{n(n-4)(n-5)}{6}}$$ The above formula $(N_0)$ is valid for polygon having $n$ no. of the sides such that $ \ \ \color{blue}{n\geq 6}$

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    $\begingroup$ Just mentioning that $N_0$ simplifies to $\dfrac{n(n-4)(n-5)}{6}$, which supports your $n \ge 6$ requirement. $\endgroup$ – steven gregory Sep 22 '15 at 13:54
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    $\begingroup$ You are absolutely correct $\endgroup$ – Harish Chandra Rajpoot Sep 22 '15 at 13:56
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    $\begingroup$ Can you elaborate a bit more on how you got the number of triangle's having one side common with the polygon as $$N_1=(n-4)n$$ $\endgroup$ – Freelancer Nov 16 '15 at 5:18
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    $\begingroup$ @Freelancer you have $n$ choice of sides. Assume you pick a side $AB$. Then, you have two less points to choose from for the third vertex. Also, the two sides that are on the right and left of $AB$ are not to be picked, for else the triangle would share two sides with the polygon. Thus, those are two less points to choose from, and you have $n-4$. Multiply the choices, and you are done. $\endgroup$ – Hasan Saad Sep 14 '16 at 8:33
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total no of triangles formed by joining vertices of n-sided polygon $$= \frac{n(n-1)(n-2)}{6}$$ i.e. selection of 3 points from n points = n(C)3 $\implies$ can also be written as sum of no of triangles formed in the following three cases,

1) no of triangles with only one side common with polygon, if we take any one side of a n-sided polygon and join its vertices to the remaining vertices, except the vertices adjacent to vertices of the line taken above, we get triangles with only one side as common i.e. for 1 side we get (n-4) triangles $\implies$ n(n-4) triangles for n sides. case I
2) no of triangles with two sides common, if we take any one side of a n-sided polygon join its vertex with its opposite vertex required triangle is formed. Hence no of triangles= n case II

3) triangles with no side common $$= \text{total - (Case I + Case II)}$$ $$=\left[\frac{n(n-1)(n-2)}{6}\right]-\left[n(n-4) + n\right]$$ $$=\frac{n(n-4)(n-5)}{6}$$

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  • $\begingroup$ Is it not just $ ^{n}C_3?$ ..and why so many views? $\endgroup$ – Narasimham Feb 4 '17 at 9:02

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