You cannot.
First: 0 to 9 is not one digit but 10 digits.
Second: You cannot change a numeric digit to a letter.
You cannot.
First: 0 to 9 is not one digit but 10 digits.
Second: You cannot change a numeric digit to a letter.
You cannot.
First: 0 to 9 is not one digit but 10 digits.
Second: You cannot change a numeric digit to a letter.
You cannot.
First: 0 to 9 is not one digit but 10 digits.
Second: You cannot change a numeric digit to a letter.
There are no 3 digit numbers between 0 and 9 because 0 and 9 are 1 digit numbers.
Look at the fourth digit. If this is 6 or more, increase preceding digit by 1. If the preceding digit is 9, change the 9 to 0 and increase its preceding digit by 1. The result in this instance is 1.000.
No. A digit is a single number from 0 to 9
a digit is any number from 0 to 9
9000. The first digit can be any one of 1-9 (9 choices). The second digit can be any one of 0-9 (10 choices). The third digit can be any one of 0-9 (10 choices). The fourth digit can be any one of 0-9 (10 choices). So, in all, 9*10*10*10 = 9000
letter -> [a-zA-Z] digit -> [0-9] identifier -> letter|_(letter|digit|_)
There are no 3 digit numbers between 0 and 9 because 0 and 9 are 1 digit numbers.
Look at the fourth digit. If this is 6 or more, increase preceding digit by 1. If the preceding digit is 9, change the 9 to 0 and increase its preceding digit by 1. The result in this instance is 1.000.
These numbers are selections from the numbers from 100 to 999. That's 9 choices for the first digit. Each time, the second digit has 9 choices (0 to 9 excluding the hundreds digit), and then the last digit has 8 choices. Total is then 9x9x8 = 648
No. A digit is a single number from 0 to 9
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9 excluding the two previous digits
a digit is any number from 0 to 9
9000. The first digit can be any one of 1-9 (9 choices). The second digit can be any one of 0-9 (10 choices). The third digit can be any one of 0-9 (10 choices). The fourth digit can be any one of 0-9 (10 choices). So, in all, 9*10*10*10 = 9000
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
If a leading zero is allowed... The first digit can be 0-9, ten possibilities. The second digit can be 0-9, ten possibilities. The third digit can be 0-9, ten possibilities. The fourth digit can be 3-9, 7 possibilities. So there are 10x10x10x7 = 7000 such codes.