A guy need to make a 8-digit code. The first 3 digits have to be different alphabets. The last 5 digits have to be one of the numbers from 1 to 9. The numbers can repeat. In how many ways can he make the 8-digit code

Sure, let's break it down:

1. For the first 3 digits, which are alphabets and must be different, we have 26 options (letters from A to Z) for the first digit, 25 options for the second digit (since it can't be the same as the first), and 24 options for the third digit (since it can't be the same as the first two). So the total number of possibilities for the first 3 digits is 26 * 25 * 24.

2. For the last 5 digits, which are numbers from 1 to 9 and can repeat, we have 9 options for each digit (since there are 9 numbers to choose from). So the total number of possibilities for the last 5 digits is 9 * 9 * 9 * 9 * 9.

3. To find the total number of ways he can make the 8-digit code, we multiply the number of possibilities for the first 3 digits by the number of possibilities for the last 5 digits. So the total number of ways is (26 * 25 * 24) * (9 * 9 * 9 * 9 * 9).

So, the guy can make the 8-digit code in (26 * 25 * 24) * (9 * 9 * 9 * 9 * 9) ways.