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10 x 9 x 8 = 720 different "permutations"

Q: How many 3 digit codes are possible with out repeating a number using 0-9 only?

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10234567

987654321

4321

There are twelve possible solutions using the rule you stated.

The smallest possible 9 digit number (assuming negatives are not allowed) is 102,345,678 in which 5 is in the thousands' place.

Related questions

987654321

10234567

9,876,543

4321

There are twelve possible solutions using the rule you stated.

The smallest possible 9 digit number (assuming negatives are not allowed) is 102,345,678 in which 5 is in the thousands' place.

Assuming whole numbers and a leading zero does not count: 10,234,567

There are many possible solutions. One such is 132486970

The number of 3-digit numbers with no repeated digits is simply 10x9x8 = 720, if you allow, for example, 012 as a 3-digit number. There are 10 digits, any of which might be the first digit. The second digit can be any digit except the digit that was used for the first digit, leaving 9 possibilities. The third digit then has 8 possibilities, since it can't be the same as the first or second digit. The actual number of possible area codes will be lower, because there are additional restrictions on the number combinations for a valid area code. For example, in North America (USA, Canada, etc.), the first digit of an area code cannot be 0 or 1 and the middle digit cannot be 9.

The least possible integer is -98765432. The least possible positive integer is 10234567.

If you don't want any odd digit, and don't want to repeat a digit, start with the largest possible digit, in this case 8, and keep adding the largest digit every time, until you run out of possibilities: 86420.

98765432