Q: What are all possible three digit numbers using numbers 0-1-2-3-4-5-6-7-8-9?

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There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)

a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.

Let's solve it a step at a time. For the first digit, how many choices do you have? 9 You can choose 1..9 but not 0, so that's nine choices for the most significant digit. For the second digit, how many choices do you have? 10 It can be 0..9. For the third digit, you also have 10 choices. Choosing one digit doesn't limit your choices for other digits and mirrored numbers (e.g. 123 and 321) are different, so all choices make a unique number. So the total is the product of our three choices: 9x10x10

6 possible 3 digit combonations

There are no three didgit numbers but there are 63 three digit numbers.

Related questions

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.

There are 900 three digit numbers. (99 - 1000) (# of possible numbers in the first position = 9) (# of possible numbers in the second position = 10) (# of possible numbers in the third position = 10) 9 *10 *10 = 900

102 = 100 which is the first possible three digit number that is a perfect square. 312 = 961 which is the last possible three digit number that is a perfect square. So there are 22 three digit positive numbers that are perfect squares.

450

There are 900 possible three-digit numbers not beginning with 0. (Note, however, that this question does not accurately describe the restrictions on numbers that can be used as area codes.)

Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.

a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.

None. The only way for it to be possible would be 3 zeros which is not considered a 3 digit numbers.

Let's solve it a step at a time. For the first digit, how many choices do you have? 9 You can choose 1..9 but not 0, so that's nine choices for the most significant digit. For the second digit, how many choices do you have? 10 It can be 0..9. For the third digit, you also have 10 choices. Choosing one digit doesn't limit your choices for other digits and mirrored numbers (e.g. 123 and 321) are different, so all choices make a unique number. So the total is the product of our three choices: 9x10x10

It is possible to create a 3-digit number, without repeated digits so the probability is 1.

6 possible 3 digit combonations

There are 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.