The game 504 is actually many games. You assemble each one by selecting one of nine modules for the top, middle, and bottom of the rules, which flip around separately to accommodate the selection.

You may think that there should be $9^3=729$ games, because for each of the three sections of the rules you have nine options. However, you cannot pick the same module for more than one section.

Considering that, you may adjust to think that there should be ${9 \choose 3} = \frac{9!}{3! \cdot (9-3)!} = 84$ games. You're selecting three modules out of a total of nine, it sounds like a combination problem. However, the order here matters. It matters if module 9 is the top or the bottom of the rules, as they describe different aspects of the game.

Thus, there are indeed $\frac{9!}{(9-3)!} = 9\cdot 8 \cdot 7 = 504$ games. This is a permutation problem. For the first, you have nine options. The second, you have eight remaining choices, and seven for the third.