First digit = 0 or 1 or 2 ... or 9 ie 10 options
Second digit = 0 or 1 or 2 ... or 9 ie 10 options
Third digit = 0 or 1 or 2 ... or 9 ie 10 options
Fourth digit = 0 or 1 or 2 ... or 9 ie 10 options
In all, 10*10*10*10 = 10000 options (combinations).
every number from 0000 to 9999
there are 10000
There are 38760 combinations.
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
9
Passwords are technically permutations, not combinations. There are 104 = 10000 of them and I regret that I do not have the time to list them. They are all the numbers from 0000 to 9999.
every number from 0000 to 9999
there are 10000
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.
There are 38760 combinations.
If every number can be used as many times as you like, there are 104 = 10000 different combinations. If each number can only be used once, there are 9!/(9 - 4)! = 5040 combinations.
There are 1140 five digit combinations between numbers 1 and 20.
There are 840 4-digit combinations without repeating any digit in the combinations.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
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.
You would get 4!/2! = 12 combinations.