There are 10 possible digits that can be used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. If you can only use each digit once, there are 10 choices for the first digit, 9 remaining choices for the second digit, and 8 remaining choices for the final digit - 10 x 9 x 8 = 720 combinations. If you can use each digit repeatedly, you have 10 choices for each digit - 10 x 10 x 10 = 1000 combinations.
10!/3! = 604800 different combinations.
Through the magic of perms and coms the answer is 729
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
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.
There are 360 of them.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
66
15
10,000
10!/3! = 604800 different combinations.
56 combinations. :)
Through the magic of perms and coms the answer is 729
9!/6!, if the six different orders of any 3 digits are considered distinct combinations.
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.
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.
There are 360 of them.
9