How many different numbers can be formed by rearranging the four digits in 2013?
Since each of the digits 0, 1, 2, and 3 is unique you can have
4! (4 factorial) possible combinations:
4! = 4x3x2x1 = 24 different ways to arrange the 4 digits
Of course any number starting with zero would usually be
truncated to 3 digits.
In that case you would omit all those arrangements from the list
of four-digit numbers leaving you with 3x3x2x1 = 18
possibilities
1) 1023
2) 1032
3) 1203
4) 1230
5) 1302
6) 1320
7) 2013
8) 2031
9) 2103
10) 2130
11) 2301
12) 2310
13) 3012
14) 3021
15) 3102
16) 3120
17) 3201
18) 3210