This question requires a combination of number theory and combinatorics. You are asked to find the number of three-digit even numbers that comply with a specific condition. In this set, 9 only ever follows 7 and 7 only ever precedes 9.

To help us simplify, we should remember that even numbers must end in 0, 2, 4, 6, or 8. Given the restriction of this problem, 79 cannot be the last two digits, because it`s not even. Therefore, the two positions where 79 or 97 can occur are the hundreds-tens or tens-ones (since this gives an even result).

With these rules in mind, let`s look at the options:

Option 1, 120 - This is too low. With two possible places for the 79 combination and other numbers to choose from, we`d definitely have more options.

Option 2, 210 - This is still likely too low for the same reasons as above.

Option 3, 365 - This seems too high - remember, we`re only looking for numbers with a specific set of digits, not all possible three-digit numbers.

Option 4, 405 - This is the correct answer. A quick way to calculate it