Since the first three digits are 537 (the exchange), we have 7-3=4 digits left. In each of those digits, there are 10 possible numbers, 0 to 9. As there are 10 numbers possible for the first digit, there are 10 numbers for each number for the second digit, and so on. For example, if the first digit was 1, we would have 1___ (3 blanks) with each blank having 10 possibilities. As there are 10 possibilities for the second number if the number 1 is there, there are 10 possibilities for the second digit for each number possible for the digit, getting 10*10=100 possible combinations with only two numbers. The same applies to the third and fourth blank, as if the second number is 2, there are 100 possibilities for 12__ (2 blanks) to be a possibility, and there are 10 blanks for each possible combination of the first two numbers, getting 10*10*10=10^3=1,000 possible combinations. Repeat the process for the fourth digit, and you end up with 10,000 possible combinations.
