It is so easy!!!
How many numbers are there between 100 and 1000 in which all digits are distinct ?
This problem is a part of Classical Combinatorics
The answer is 648.
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We have to form all possible 3-digits numbers with distinct digits.
We can't have 0 on the hundred 's place. So,there nine ways of filling the hundred's place.
Now,9 digits are left including 0. So,ten's place can be filled with any of the remaining 9 digits in 9 ways.
Now ,the unit's place can be filled with in any of the remaining 8 digits. So,there are 8 ways of the filling the unit's place.
Hence,The Total number of required numbers= 9 * 9 * 8 = 648 . (Here star(*) refers to multiplication.)