Example 2: How many odd numbers with all different digits are there between 1000 and 9999?
In the numbers between 1000 and 9999, all are ordered arrangements of 4 digits. Therefore, to find the number of certain arrangements, we need to make 4 choices: the units, tens, hundreds, and thousands digits. Since the number we want is odd, the units digit can be any one of 1, 3, 5, 7, 9. The tens and hundreds digits can be any one of 0, 1, 2, ..., 9, and the thousands digit can be any one of 1, 2, ..., 9. Thus, there are 5 choices for the units digit. Since the four digits must be different, after the units digit is chosen, there are only 8 choices left for the thousands digit. After these two digits are chosen, there are 8 choices left for the hundreds digit. And after these three digits are chosen, there are 7 choices left for the tens digit. Therefore, by the multiplication principle, the answer to this problem is \(5 \times 8 \times 8 \times 7 = 2240\) kinds.