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Leaving Zero out, we would operate in a nine digit numbersystem.

9*9*9*9, a total of 6561 different numbers.

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If Zero is included, then the total possible numbers would be:

10*10*10*10, a toatal of 10.000 different numbers.

you can easily calculate the answer for any given numbersystem.

Say if you want 5 digits, then just do 9 to the power of 5 With a calculator.

You can calculate this answer with any numbersystem this way.

For a 16 digit number system, 4 digits, the answer would be 16 to the power of 4.

Math is accurate and never lies.

Q: If using numbers 1 through 9 how many different 4 digit combinations can be made?

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10!/3! = 604800 different combinations.

Through the magic of perms and coms the answer is 729

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.

There are 360 of them.

Related questions

Number of 7 digit combinations out of the 10 one-digit numbers = 120.

66

10,000

15

10!/3! = 604800 different combinations.

56 combinations. :)

Through the magic of perms and coms the answer is 729

9!/6!, if the six different orders of any 3 digits are considered distinct combinations.

If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.

There are 360 of them.

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.

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