How many numbers greater than million can be formed with the digits 2,3,0,3,4,2 and 3 if repetition of digits is not allowed?

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.`