Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then:
If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers
If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers.
If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then:
If repeats are not allowed there are 4 possible groups
If repeats are allowed there are 20 possible groups.
There are twelve possible solutions using the rule you stated.
1
-8
10,000 combinations.
There are 15180 combinations.
You have 2 options for the first digit, 2 options for the second digit, etc. ... In total, that gives you 210 combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
216
There are 840 4-digit combinations without repeating any digit in the combinations.
There are 5,040 combinations.
There are 210 4 digit combinations and 5040 different 4 digit codes.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
Assuming no repeated digits, lowest first, 20; in any order 120; Allowing repeated digits: 216
There are twelve possible solutions using the rule you stated.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
1