How many different numbers can be formed by rearranging the four digits in 2013?

Answer 1

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

Answer 2

3 options for the first digit (since I am assuming the number won't start with a zero), 3 for the second, 2 for the third, 1 for the fourth. Multiply everything together.

Answer 3

how make the answer im don/now