The sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once is

To solve this problem, we need to understand that each of the digits 3, 5, 5, and 6 will appear in each of the thousands, hundreds, tens, and ones places the same number of times when we list all possible 4-digit numbers.

1. There are 4!/(2!) = 12 different 4-digit numbers that can be formed using the digits 3, 5, 5, and 6. The division by 2! is because the digit 5 is repeated.

2. Each digit will appear in the thousands, hundreds, tens, and ones places 12/4 = 3 times.

3. So, the sum of all these numbers will be the sum of the 4-digit numbers where each digit appears in each place 3 times.

4. The sum of the numbers formed by the digits 3, 5, 5, and 6 in each place is 3*(3+5+5+6) = 57.

5. Therefore, the sum of all the possible 4-digit numbers is 57*(10^3 + 10^2 + 10^1 + 10^0) = 57*(1111) = 63327.