After doing some research, I found out that the number of diagonals of an *n*-sided polygon = $\frac{n(n-3)}{2}$.

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This formula works, of course, but the question is one of those in my textbook designated to be solved using *combinations*. Therefore, if a similar question came up in a test, I could only imagine the correct working out involving combinations.

I can imagine that said formula could be a simplification of $\binom{n}{r}$ = $\frac{n!}{r!(n-r)!}$. However, I"m unsure of the specifics.

combinatorics binomial-coefficients combinations

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asked Aug 19 "16 at 1:58

Mad BannersMad Banners

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By a happy coincidence an $n$-sided polygon also has $n$ vertices. A diagonal joins two vertices, which can be done in $\binom n2$ ways. However, this count includes the $n$ sides, so subtract $n$ to get the number of diagonals:$$\binom n2-n=\frac{n(n-1)}2-\frac{2n}2=\frac{n(n-3)}2$$

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answered Aug 19 "16 at 2:06

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Each diagonal can be thought of as a pair of vertices (where order doesn"t matter), of which there are ${n\choose2} = \frac{n(n-1)}{2}$. This also counts each side, though, so you subtract $n$ to get $\frac{n(n-3)}{2}$.

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answered Aug 19 "16 at 2:06

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