Through the magic of perms and coms the answer is 729
10!/3! = 604800 different combinations.
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.
9
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
66
10,000
15
10!/3! = 604800 different combinations.
56 combinations. :)
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.
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.
9
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.