you could make a probability tree if you could be bothered
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
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.
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
Well if repeats are allowed, then unlimited. If you can repeat the numbers then the list would go on forever.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
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.
In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.
120
There are 840 4-digit combinations without repeating any digit in the combinations.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
This question needs clarificatioh. There are 4 one digit number combinations, 16 two digit combinations, ... 4 raised to the n power for n digit combinations.
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
We are not told anything about repeats so I will assume you can repeat a digit. For example, you can have 111111 The first digits has 50 choices so does the second and the third..etc so we have 50^6 combinations.