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.