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10!/3! = 604800 different combinations.

Q: How many 7 digit combinations can you have if you can only use the numbers 0 through 9 once each?

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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.

The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .

it's 610, which is 60,466,176. Since each digit spot could be anything 0-9 there are 10 numbers to try for each digit, making it 610.

There are 10 possible digits that can be used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. If you can only use each digit once, there are 10 choices for the first digit, 9 remaining choices for the second digit, and 8 remaining choices for the final digit - 10 x 9 x 8 = 720 combinations. If you can use each digit repeatedly, you have 10 choices for each digit - 10 x 10 x 10 = 1000 combinations.

99. Each of the numbers in the list can only be used by itself otherwise you would get a 4-digit (or longer) number.

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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.

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

120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.

The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .

If all numbers can be used as many times as wanted then there are 109 = one billion combinations. If each number can be used only once, there are 10!/(10-9)! = 10!/1! = 10! = 3628800 combinations. * * * * * Clearly answered by someone who does not know the difference between PERMUTATIONS and COMBINATIONS. The combination 123456789 is the same as the combination 213456789 etc. All in all, therefore, there are only ten combinations which use each digit at most once.

It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.

it's 610, which is 60,466,176. Since each digit spot could be anything 0-9 there are 10 numbers to try for each digit, making it 610.

There are 10 possible digits that can be used: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. If you can only use each digit once, there are 10 choices for the first digit, 9 remaining choices for the second digit, and 8 remaining choices for the final digit - 10 x 9 x 8 = 720 combinations. If you can use each digit repeatedly, you have 10 choices for each digit - 10 x 10 x 10 = 1000 combinations.

99. Each of the numbers in the list can only be used by itself otherwise you would get a 4-digit (or longer) number.

There are 10,000 possible combinations, if each number can be used more than once.

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 ten combinations: one each where one of the ten digits, 0-9, is excluded.