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# The number of diagonals in a octagon will be

$\begin{array}{1 1}(A)\;28\\(B)\;20\\(C)\;10\\(D)\;16\end{array}$

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• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
In a octagon there are eight sides and eight points.The diagonal will be formed by joining any two points except the sides.
$\therefore$ Required number of ways =$8C_2-8$
$8C_2=\large\frac{8!}{2!6!}$
$\Rightarrow \large\frac{8\times 7\times 6!}{2\times 6!}$
$\Rightarrow 28$
$8C_2-8=28-8$
$\Rightarrow 20$
Hence (B) is the correct answer.