Final answer:
The total number of eight-digit numbers with the first four digits being odd and the last four being even is 390625. If repetition is not allowed, the total number of such eight-digit numbers is 14400.
Step-by-step explanation:
The question revolves around creating eight-digit numbers with certain constraints. For the first part, we want the first four digits to be odd and the last four to be even. Since there are five odd digits (1, 3, 5, 7, 9) and five even digits (0, 2, 4, 6, 8), each position in the first half can be filled by any of the five odd digits, and similarly, each position in the second half can be filled by any of the five even digits.
Therefore, the total number of such numbers is 5^4 * 5^4, which equals 625*625 or 390625.
For the second part, if repetition is not allowed, the first digit can be any of the five odd digits. The second digit can then be any of the remaining four odd digits, and so on. The same rule applies for the even digits in the second half of the number.
Thus, the total number of non-repetitive eight-digit numbers is 5! * 5!, which equals 120*120 or 14400.