Given : 0 to 9 digits (10 digits)
To find : 4-digit numbers
First, thousand’s place on be filled by 9 ways from any one of the 1 to 9 digits (v ‘0’ (zero) is not placed in thousand’s place).
Then the remaining three places can be filled with the remaining 9 digits in 9P3 ways.
∴ Required number of numbers = 9 x 9P3
= 9 x (9 x 8 x 7) = 4536