How many numbers greater than 10 lakhs will be formed from 2, 3, 0, 3, 4, 2, 3?

All seven digits are included in any number that is greater than one million.
The number of arrangements of seven digits $=\frac{7!}{2!\times 3!}$
$=\frac{7\times 6\times 5\times 4\times 3!}{2!\times 3!}$
$=7\times 6\times 5\times \mathrm{ }2$
$=420$
Now, there are 6 digits, with 0 set at the millionth position, 2, 3, and 3 twice, and the other digits being different.
These 6 digits can be arranged = $\frac{6!}{2!\times 3!}$
$=\frac{6\times 5\times 4\times 3!}{2!\times 3!}$
$=6\times 5\times 2$
$=60$ ways
Hence, the number of required numbers $=420-60=\mathbf{360}$