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# In a polygon no three diagonals are concurrent.If the total number of points of intersection of diagonals interior to the polygon be 70,then the number of diagonals of the polygon is

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

Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
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.