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you could make a probability tree if you could be bothered

Q: How do you generate a list of five digit combinations of only the numbers 1 2 3 4 and 5 with out repeats?

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Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.

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 38760 combinations.

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.

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.

Related questions

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.

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 38760 combinations.

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.

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.

There are 1140 five digit combinations between numbers 1 and 20.

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.