Let's solve it a step at a time.
For the first digit, how many choices do you have? 9
You can choose 1..9 but not 0, so that's nine choices for the most significant digit.
For the second digit, how many choices do you have? 10
It can be 0..9.
For the third digit, you also have 10 choices.
Choosing one digit doesn't limit your choices for other digits and mirrored numbers (e.g. 123 and 321) are different, so all choices make a unique number.
So the total is the product of our three choices: 9x10x10
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There are 76 such numbers. Eight more if you allow numbers to start with 0.
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated