If the numbers are not allowed to repeat themselves then 362,880.
* * * * *
That is the number of permutations, not combinations.
In a combination the order of the digits does not matter. So there is only one 9-digit combination.
I did 1×2×3×4×5×6×7×8×9 and got 362,880
362,880
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.
The answer is 10C4 = 10!/[4!*6!] = 210
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Through the magic of perms and coms the answer is 729
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
It would be from 1111111 to 9999999, and because 9999999 - 1111111 = 8888888, the answer is 8888888.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
It is: 9C7 = 36
66
6 ways: 931,913,139,193,391,319
10,000
15
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.
The answer is 10C4 = 10!/[4!*6!] = 210
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.