$\begin{array}{1 1}(A)\;20\\(B)\;28\\(C)\;8\\(D)\;\text{None of these}\end{array} $

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

A selection of four vertices of the polygon gives an interior intersection

$\therefore$ The number of sides =n$\Rightarrow nC_4=70$

$n(n-1)(n-2)(n-3)=24\times 70$

$\Rightarrow 8\times 7\times 6\times 5$

$\Rightarrow n=8$

$\therefore$ The number of diagonals =$8C_2-8$

$8C_2=\large\frac{8!}{2!\times 6!}$

$\Rightarrow \large\frac{8\times 7\times 6!}{2\times 6!}$$=28$

$8C_2-8=28-8$

$\Rightarrow 20$

Hence (A) is the correct answer.

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