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Q: What are all the possible 5 digit codes using the numbers 0-9?

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There are 210 4 digit combinations and 5040 different 4 digit codes.

The possible 4 digit codes using the numbers 0-9 are every number between 0 and 9999. For numbers that have less than 4 digits, just precede the number with 0's. 10,000 possibilities

Any 4 from 10 in any order = 10 x 9 x 8 x 7 = 5040

There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.

Total number of possible 3-digit numbers = 9!x10!10!

1000

1,000 of them. The list of possibilities will look exactly like the counting numbers from 000 to 999 .

Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.

103 = 1000 in all.

10,000

From {1, 2, 3, 4} there are two prime numbers (2, 3} which can go in the hundreds position, which once chosen allows 3 possible choices for the tens digit and a further 2 possible choices for the units digit, giving 2 x 3 x 2 = 12 possible numbers.

6 possible 3 digit combonations

9999999999 100000000

Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.

it depends on which 8 numbers your talking about

-123456786

There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.

290

This is not possible, since there are only five single digit odd numbers, which are 1, 3, 5, 7 and 9.

,888

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.

There are twelve possible solutions using the rule you stated... 13,14,17,31,34,37,41,43,47,71,73 & 74

Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991

It depends on how many digits in an area code. North America (USA, Canada, etc.) uses a uniform system of 3-digit area codes, but many other countries do not. Assuming a 3-digit area code and no other restrictions other than "first digit not zero," that leaves 900 possible area codes.

If a digit can be repeated there are 5 x 5 x 5 = 125 possible numbers If a digit cannot be repeated there are 5 x 4 x 3 = 60 possible numbers.