99. Each of the numbers in the list can only be used by itself otherwise you would get a 4-digit (or longer) number.
5
There would be 9*9*9*9 or 6561 combinations.
-8
Exactly 3,628,800, or 10!.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
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.
99. Each of the numbers in the list can only be used by itself otherwise you would get a 4-digit (or longer) number.
There are 1140 five digit combinations between numbers 1 and 20.
You would get 4!/2! = 12 combinations.
There are 840 4-digit combinations without repeating any digit in the combinations.
7
56 combinations. :)
Any 6 from 51 = 18,009,460 combinations
5
If all numbers can be used as many times as wanted then there are 109 = one billion combinations. If each number can be used only once, there are 10!/(10-9)! = 10!/1! = 10! = 3628800 combinations. * * * * * Clearly answered by someone who does not know the difference between PERMUTATIONS and COMBINATIONS. The combination 123456789 is the same as the combination 213456789 etc. All in all, therefore, there are only ten combinations which use each digit at most once.
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.