Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
it depends on which 8 numbers your talking about
Exactly 3,628,800, or 10!.
16
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
56 combinations. :)
90
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
7
it depends on which 8 numbers your talking about
Exactly 3,628,800, or 10!.
We can use 36 characters for each of the slots in the combination. Therefore, we have 36^11 possible combinations, or 131,621,703,842,267,136 combinations.
Without restrictions, it was would numbers 000-000-0000 through 999-999-9999. So that would be 9,999,999,999 + 1 = 10 billion different 10-digit phone numbers. Ex: If there existed single digit phone numbers, there would be 10, because the digits are 0 through 9. If there existed only double digit phone numbers, then it would be 00 through 99 which would be 100 total two-digit numbers. Therefore the total possible combinations for an X digit phone number would be: 10^X