Final answer:
In this question, we determine the number of six-digit integers with no repeated digits and with repeated digits. We also consider the extra condition of the integers being even or divisible by 5.
Step-by-step explanation:
In part (a), the number of six-digit integers with no repeated digits can be found using the concept of permutations. Since there are no leading zeros, the first digit can be chosen from 9 options (1 to 9), the second digit can be chosen from 9 options (any digit except the one already chosen), and so on. Therefore, the number of such integers is 9 * 9 * 8 * 7 * 6 * 5 = 136,080.
In part (b), the number of six-digit integers with repeated digits can be found by using the concept of combinations. Each digit can be chosen from 10 options (0 to 9), and there are 6 digits. Therefore, the number of such integers is 10^6 = 1,000,000.
In both parts (a) and (b), if the six-digit integer needs to be even, the last digit can only be chosen from the set {0, 2, 4, 6, 8}. Therefore, the number of even six-digit integers with no repeated digits is 9 * 9 * 8 * 7 * 6 * 5 * 5 = 68,400, and the number of even six-digit integers with repeated digits is 10^5 * 5 = 500,000.
If the six-digit integer needs to be divisible by 5, the last digit can only be 0 or 5. Therefore, the number of six-digit integers divisible by 5 with no repeated digits is 9 * 9 * 8 * 7 * 6 * 2 = 48,384, and the number of six-digit integers divisible by 5 with repeated digits is 10^5 * 2 = 200,000.