How many 4-digit numbers are there, when a digit may be repeated any number of times?

Explanation

Clearly, 0 cannot be placed at the thousand's place.
So, this place can be filled with any digit from 1 to 9.
Thus, there are 9 ways of filling the thousand's place.
Since repetition of digits is allowed, each one of the remaining 3 places can be filled in 10 ways, i.e., with any digit from 0 to 9.
So, the required number of numbers `=(9xx10^(3))=9000.`