Each Student at a certain university is given a 6-digit code (such as 123789 or 001122)
(a) how many different codes are there?
(b) how many codes read the same forward and backward?
(c) how many codes contain only odd digits?
(d) how many codes contain at least one even digit?
To calculate the number of different codes, we need to determine the possibilities for each digit of the code.
(a) The code has 6 digits, and each digit can take any value from 0 to 9. Since repetition is allowed, there are 10 options for each digit. Therefore, the total number of different codes is calculated as: 10*10*10*10*10*10 = 1,000,000.
(b) To find the number of codes that read the same forward and backward (palindromic codes), we need to consider the possibilities for the first three digits since the last three digits will automatically replicate those. The first digit can be any number from 1 to 9 (excluding 0), as leading zeros are not allowed to maintain the 6-digit format. The second digit can be any number from 0 to 9, and for the third digit, we only have the same options as the first digit (1 to 9). Therefore, the number of palindromic codes is calculated as: 9*10*1 = 90.
(c) To calculate the number of codes that contain only odd digits, we analyze each digit independently. The first digit can only be an odd number, so we have 5 options (1, 3, 5, 7, 9). The same applies to the second, third, fourth, fifth, and sixth digits. Therefore, the number of codes containing only odd digits is calculated as: 5*5*5*5*5*5 = 15,625.
(d) To find the number of codes that contain at least one even digit, we can subtract the number of codes that contain only odd digits (calculated in part (c)) from the total number of codes (calculated in part (a)). Therefore, the number of codes with at least one even digit is: 1,000,000 - 15,625 = 984,375.