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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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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 =$8C_2-8$
$\Rightarrow \large\frac{8\times 7\times 6!}{2\times 6!}$
$\Rightarrow 28$
$\Rightarrow 20$
Hence (B) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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