a personal identification number consists of 3 letters followed by 3 digits (0-9). the letters x,y and z are not used and the last digit cannot be 0 or 9. how many different identification numbers are possible?

The English alphabet consists of 26 letters, but since x, y and z are not used, we'll have 23 letters to choose from. From 0 to 9, there are 10 numbers total, but there's restriction that the last digit cannot be 0 or 9, so that's 8 numbers total to choose from for the last digit.

Assuming repetition is ALLOWED:
23 * 23 * 23 * 10 * 10 * 8 = 9733600

Well, it seems like they really didn't want to use the letters x, y, and z in the identification numbers. I guess those letters felt left out!

Anyway, to calculate the number of possible identification numbers, we need to multiply the number of choices for each position.

For the first letter, we have 23 choices (26 letters minus 3 not used). For the second letter, we still have 23 choices. And for the third letter, once again, we have 23 choices.

Moving on to the digits, for each of the three positions, we have 8 choices (digits 1-8, since the last digit cannot be 0 or 9).

So, to find the total number of possible identification numbers, we multiply all these choices together:

23 choices for the first letter * 23 choices for the second letter * 23 choices for the third letter * 8 choices for the first digit * 8 choices for the second digit * 8 choices for the third digit =

23 * 23 * 23 * 8 * 8 * 8 = 8,704,768.

So, there are 8,704,768 different possible identification numbers that fit these criteria. That's a lot of numbers!

To calculate the number of different identification numbers, we need to determine the number of options for each character and then multiply them together.

1. For the first letter, we have 26 options (all the letters of the alphabet minus the three excluded letters x, y, and z).
2. For the second letter, we also have 26 options.
3. For the third letter, we have 26 options.
4. For the first digit, we have 10 options (0-9).
5. For the second digit, we have 10 options.
6. For the third digit, we have 7 options (1-7, since the last digit cannot be 0 or 9).

To calculate the total number of different identification numbers, we multiply these options together:

26 * 26 * 26 * 10 * 10 * 7 = 4,368,000

Therefore, there are 4,368,000 different possible identification numbers in this case.

To find the number of different identification numbers possible, we need to calculate how many choices we have for each character in the ID.

1. First, let's consider the three letters. As stated in the problem, the letters X, Y, and Z are not used. So, we have 23 letters available (26 letters in the alphabet minus the 3 excluded letters).

2. Now, let's move on to the three digits. We have 10 choices for each digit (0-9). However, the last digit cannot be 0 or 9, so we have 8 choices for the last digit.

To find the total number of possible identification numbers, we multiply the number of choices for each character together:

Number of letter choices: 23 * 23 * 23 = 12,167
Number of digit choices: 10 * 10 * 8 = 800

Total number of possible identification numbers: 12,167 * 800 = 9,733,600

Therefore, there are 9,733,600 different possible identification numbers.