Exactly 3,628,800, or 10!.
10!/3! = 604800 different combinations.
Through the magic of perms and coms the answer is 729
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
Exactly 3,628,800, or 10!.
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.
It is: 9C7 = 36
10,000
15
10!/3! = 604800 different combinations.
that would be 9 to the 7th power i believe.
Through the magic of perms and coms the answer is 729
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
You would get 4!/2! = 12 combinations.
There are 840 4-digit combinations without repeating any digit in the combinations.