There are 10 digits in the number 2,563,183,083. If we ignore the fact that there are three occurrences of the digit 3 and two occurrences of the digit 8, then there would be 10! = 3,628,800 distinct ways to arrange the digits.
However, we need to account for the fact that three of the digits are 3s and two of the digits are 8s. To do this, we need to divide out the number of arrangements of those duplicates.
There are 3! = 6 ways to arrange the three 3s among themselves, and 2! = 2 ways to arrange the two 8s among themselves. So, we divide the total number of arrangements by these permutations of the duplicates:
10! / (3! * 2!) = 10 x 9 x 8 x 7 x 6 x 5 x 4 = 604,800
Therefore, there are 604,800 distinct ways to arrange the digits in the number 2,563,183,083, assuming that any zero digits can be placed in any position.