$\begin{array}{1 1}(A)\;300\\(B)\;310\\(C)\;320\\(D)\;330\end{array} $

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

White marbles =6

Red marbles =5

Total no of marbles =11

No of ways in which 4 marbles can be drawn =$11C_4$

$11C_4=\large\frac{11!}{4!(11-4)!}$

$\Rightarrow \large\frac{11!}{4!7!}$

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

$\Rightarrow 330$ ways

Hence (D) is the correct answer.

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