How many different positive integers can be formed using each digit in

There are 7 digits out of which there are three 3s.
Different ways these can be arranged is 7!/3!, as three 3s can be arranged in 3! Ways we divided total by 3!
7!/3!=7*6*5*4=840.
However 840 is not in choices, and we have to use each digit once. But the numbers starting with 0 actually do not use 0 for example 0133378 is same as 133378, but without digit 0. Thus the question means 7-digit positive integers, and in this case we have to remove numbers starting with 0.
If first digit is 0, the remaining 6 can be arranged in 6!/3!=6*5*4=120
Total =840-110=720

OTHER WAY
The first digit can be taken by any of the digits except 0, so 7-1=6 ways. The remaining 6 positions can be arranged in 6!ways. —-6*6!
But there are 3 similar digits so total ways =6*6!/3!=6*6*5*4=720


B