$\begin{array}{1 1}(A)\;10\\(B)\;15\\(C)\;20\\(D)\;25\end{array} $

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

Total no of white marbles =6

Total no of red marbles =5

The no of ways of which all are white marbles =$6C_4$

All are red marbles =$5C_4$

Total no of ways =$6C_4+5C_4$

$6C_4=\large\frac{6!}{4!2!}$

$\Rightarrow \large\frac{6\times 5\times 4!}{2\times 1\times 4!}$

$\Rightarrow 15$

$5C_4=\large\frac{5!}{4!1!}$

$\Rightarrow \large\frac{5\times 4!}{1\times 4!}$

$\Rightarrow 5$

Total no of ways =15+5=20

Hence (C) is the correct answer.

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