$\begin{array}{1 1}(A)\;270\\(B)\;220\\(C)\;282\\(D)\;\text{None of these}\end{array} $

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

The number of triangles with vertices on side AB,BC,CD=$3C_1\times 4C_1\times 5C_1$

Similarly for other cases

$\therefore$ The total number of triangles =$3C_1\times 4C_1\times 5C_1+3C_1\times 4C_1\times 6C_1+3C_1\times 5C_1\times 6C_1+4C_1\times 5C_1\times 6C_1$

$3C_1=\large\frac{3!}{1!2!}$$=3$

$4C_1=\large\frac{4!}{1!3!}$$=4$

$5C_1=\large\frac{5!}{1!4!}$$=5$

$6C_1=\large\frac{6!}{1!5!}$$=6$

$\Rightarrow 3\times 4\times 5+3\times 4\times 6+3\times 5\times 6+4\times 5\times 6$

$\Rightarrow 342$

Hence (D) is the correct answer.

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