$\begin{array}{1 1}(A)\;500\\(B)\;505\\(C)\;510\\(D)\;\text{None of these}\end{array} $

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

Required number of ways =Total number of ways choosing 4 pens -number of ways of choosing 4 non-red pens

$\Rightarrow {(3+4+6)}C_4-{4+6}C_4=13C_4-10C_4$

$\Rightarrow \large\frac{13!}{4!9!}-\frac{10!}{4!6!}$

$\Rightarrow \large\frac{13\times 12\times 11\times 10\times 9!}{4\times 3\times 2\times 1\times 9!}-\frac{10\times 9\times 8 \times 7\times 6!}{4\times 3\times 2\times 1\times 6!}$

$\Rightarrow 715-210$

$\Rightarrow 505$

Hence (B) is the correct answer.

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