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# A bag contains six white marbles and five red marbles.Find the number of ways in which four marbles can be drawn from the bag if two must be white and two red

$\begin{array}{1 1}(A)\;130\\(B)\;140\\(C)\;150\\(D)\;160\end{array}$

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• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total no of white marbles =6
Total no of red marbles =5
No of ways of choosing two white and two white and two red are =$6C_2\times 5C_2$
$6C_2=\large\frac{6!}{2!\times 4!}$
$\Rightarrow \large\frac{6\times 5\times 4!}{2\times 1\times 4!}$
$\Rightarrow 15$
$5C_2=\large\frac{5!}{2!3!}$
$\Rightarrow \large\frac {5\times 4\times 3!}{2\times 3!}$
$\Rightarrow 10$
Total no of ways =$15\times 10$
$\Rightarrow 150$ way
Hence (C) is the correct answer.