The number of different numbers greater than $1000000$ that can be formed with the digits $2,3,0,3,4,2,3$ is

The correct option is B $360$

The total number of $7$ digit numbers formed using the given digits is $\frac{7!}{2!3!}$ because $2$ and $3$ are repeated.

Now eliminating the ones starting with $0$, we get

$\frac{7!}{2!3!}-\frac{6!}{2!3!}$

$=\frac{6.(6!)}{2!3!}$

$=\frac{6!}{2!}$

$=360$