How many numbers greater than a million can be formed with the digits $2,3,0,3,4,2,3$?

Any number of greater than a million will contain all the seven digits.

Now, we have to arrange these seven digits, out of which $2$ occurs twice, $3$ occurs twice and the rest are distinct.

The number of such arrangements $=\frac{7!}{2!\times 3!}=\frac{7\times 6\times 5\times 4\times 3!}{2!\times 3!}=7\times 6\times 5\times 2=420$

These arrangements also include those numbers which contain $0$ at the million’s place.

Keeping $0$ fixed at the millionth place, we have $6$ digits out of which $2$ occurs twice, $3$ occurs thrice and the rest are distinct.

These arrangements also include those numbers which contain $0$ at the million’s place.

Keeping $0$ fixed at the millionth place, we have $6$ digits out of which $2$ occurs twice, $3$ occurs thrice and the rest are distinct.

These $6$ digits can be arranged in $\frac{6!}{2!\times 3!}=\frac{6\times 5\times 4\times 3!}{2!\times 3!}=6\times 5\times 2=60$  ways

Hence, the number of required numbers $=420–60=360$.