Using the word "combinations" in the English sense (as opposed to mathematical sense the expert has used) where it often used in the mathematical sense of the word "permutations":
Assuming the hundreds digit must be at least 1 (eg 99 = 099 is not considered a three digit number), then:
9 x 10 x 10 = 900.
If you are allowed to repeat letters (A-Z) then there are 17,576 different combinations. If you can not repeat letters then there are 15,600 combinations.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
There are 210 4 digit combinations and 5040 different 4 digit codes.
9^9 = 387420489 different digit combinations.
if its not alphanumeric, 999999 variations
There are 840 4-digit combinations without repeating any digit in the combinations.
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
If you can repeat a digit, there are 27. If you can't repeat a digit, there are only 6.
There are 5,040 combinations.
There are 360 of them.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
This is a factorial problem. The first number can be any of ten digits, the second any of nine (because you can't repeat a digit), the third any of eight and the fourth any of the remaining 7 digits. 10x9x8x7=5040 combinations.
I believe the answer would be 10^9 or 1,000,000,000
0000, 0001, 0002, ...., 9997, 9998, 9999