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

Explanation

Any number greater than one million will contain all the seven digits.
Now, we have to arrange seven digits, out of which 2 occurs twice, 3 occurs thrice and the rest are distinct.
Number of such arrangements `=(7!)/((2!)xx(3!))=420.`
These arrangements will also include thsoe which contain 0 at the million's place.
Keeping 0 fixed at the million's place, the remaining 6 digits out of which 2 occurs twice, 3 occurs thrice and the rest are distinct can be arranged in `=(6!)/((2!)xx(3!))=60` ways.
Hence, the number of required numbers `=(420-60)=360.`