This question could would be best answered if you have a basic understanding of combinatorics (the study of combinations and permutations and probability) however, for something with as few possibilities as this, it is often easier to just make a list. Start with three digit combinations, then two digit combos, then one digit. Remember also that combinations are order independent ( ABC is the same thing as CBA)
3 digits
3 2 1
2 digits
3 2
3 1
2 1
1 digits
1
2
3
count those up and voila.
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.
9
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
There are 840 4-digit combinations without repeating any digit in the combinations.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
There are 5,040 combinations.
216
There are 210 4 digit combinations and 5040 different 4 digit codes.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
There are twelve possible solutions using the rule you stated.
6 - 123, 132, 213, 231, 312, and 321.
Well, honey, there are 720 three-digit combinations using the numbers 0-9 without repetition. So, if you're looking for a list, you better grab a pen and start writing because I ain't got time to list them all out for you! Just remember, math is your friend, darling.