$\begin{array}{1 1}(A)\;3\\(B)\;36\\(C)\;66\\(D)\;108\end{array} $

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

Total no. of balls in urn A=3

Total no. of balls in urn B=9

The number of ways in which two balls from urn A and two balls from urn B can be selected =$3C_2\times 9C_2$

$3C_2=\large\frac{3!}{2!1!}$$=3$

$9C_2=\large\frac{9!}{2!7!}=\frac{9\times 8\times 7!}{2!\times 7!}$$=36$

$3C_2\times 9C_2=3\times 36$

$\Rightarrow 108$

Hence (D) is the correct answer.

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