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The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the octagon is

$\begin{array}{1 1}(A)\;16\\(B)\;28\\(C)\;56\\(D)\;70\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Required number of triangles =total number of triangles-number of triangles having two sides common -number of triangles having one side common
$\Rightarrow 8C_3-8-8\times 4$
$\Rightarrow \large\frac{8!}{3!5!}$$-8-8\times 4$
$\Rightarrow \large\frac{8\times 7\times 6\times 5!}{3\times 2\times5!}$$-8-32$
$\Rightarrow 56-8-32$
$\Rightarrow 56-40$
$\Rightarrow 16$
Hence (A) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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