$\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.

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