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We have to find the numbers of diagonals in the n-sided polygon.

The number of line segments obtained by joining the vertices of a n sided polygon taken two points at a time.

Now, applying the formula and using permutation as below stated.

The number of ways of selecting 2 points at a time from n number of points is given as \[^{\text{n}}{{\text{C}}_{\text{2}}}\]

As we have \[^{\text{n}}{{\text{C}}_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

So we have

\[^{\text{n}}{{\text{C}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{2!}}\left( {{\text{n - 2}}} \right){\text{!}}}}\]

On simplifying we get,

\[{ \Rightarrow ^{\text{n}}}{{\text{C}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{n(n - 1)(n - 2)!}}}}{{{\text{2!}}\left( {{\text{n - 2}}} \right){\text{!}}}}\]

\[ \Rightarrow \dfrac{{{\text{n(n - 1)}}}}{{\text{2}}}\]

Hence, out of the total selections here n are the sides of the polygon so subtracting that from the total selections, we get,

\[ \Rightarrow \dfrac{{{\text{n(n - 1)}}}}{{\text{2}}} - n\]

On simplifying we get,

\[ \Rightarrow \dfrac{{{\text{n(n - 1) - 2n}}}}{{\text{2}}}\]

On taking n common from both the terms we get,

\[

\Rightarrow \dfrac{{{\text{n(n - 1 - 2)}}}}{{\text{2}}} \\

\Rightarrow \dfrac{{{\text{n(n - 3)}}}}{{\text{2}}} \\

\]

Hence , there are total \[\dfrac{{{\text{n(n - 3)}}}}{{\text{2}}}\] number of diagonals in an n sided polygon.