To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
6
Only one.
10
Six combinations: 123, 132, 213, 231, 312, 321
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
If repeats are allowed than an infinite number of combinations is possible.
10,000.
45
5040, assuming none of the digits are the same. (Assuming they're not, there's 5040 unique combinations you can make out of 7 digits).
6
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.
Only one.
Assuming no duplication or repetition, 720
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
10
You Can Create 999 Number combinations
32C3 = 4960