The answer depends on what is implied by "such".
There are 900 three-digit numbers.
1344
6 if all digits are different, 27 otherwise.
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
The numbers from 1000 to 9999 are all 4 digits, as are -1000 to -9999. Therefore there are 18,000 four-digit numbers in all.
There are 900 three-digit numbers.
1344
6 if all digits are different, 27 otherwise.
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
2893 digits in all, made up as follows: 9 1-digit numbers (1 to 9) = 1*9 = 9 digits 90 2-digit numbers (10 to 99) = 2*90 = 180 digits 900 3-digit numbers (100 to 999) = 3*900 = 270 digit 1 4-digit number (1000) = 4*1 = 4 digits
The numbers from 1000 to 9999 are all 4 digits, as are -1000 to -9999. Therefore there are 18,000 four-digit numbers in all.
The numbers from 1000 to 9999 are all 4 digits, as are -1000 to -9999. Therefore there are 18,000 four-digit numbers in all.
There are 90 positive numbers with two digits. This is because the range of two-digit numbers is from 10 to 99 (inclusive), and there are 90 numbers within this range.
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.
18, if leading 0s are disallowed.
There are only five distinct odd digits.
Zero; all six digit numbers have six digits. Dude.. i m asking 6 digit numbers, containing only 4 different digits..!! e.g.:123412.. with only 2 digits repeated..!! My guess is: 10*9*8*7*4*3 but I'm also guessing that is wrong- sorry!