$\begin{array}{1 1}(A)\;30\\(B)\;60\\(C)\;90\\(D)\;120\end{array} $

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- $n!=n(n-1)(n-2)(n-3).......(3)(2)(1)$

Total no. of digits =6

Total no. =112233

From 112233,the number of 6 digit that can be formed are

$\Rightarrow \large\frac{6!}{2!2!2!}$

$\Rightarrow \large\frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 2 \times 2}$

$\Rightarrow 90$

Hence (C) is the correct answer.

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