This is a question dealing with combinations and permutations. It sounds like it's a two-part question; first you have to figure out how many different possible lineups there are. Then, you have to figure out how many seasons that will fill up. The first part is the only really hard part.

So you have 12 players, and there are 9 positions. Let's just look at how many different ways there are to line up those 12 players into 9 spots. For the first position, you have your pick of all 12 players, right? So we've already got 12 options just in the first slot. For the second slot, you'll only have 11 options, because one player has already been used. And for the third, you'll only have 10 options, and so on down the line until the 9th slot, where you'll have 4 people left to choose from.

Now here's the catch: for each of the 11 possible second-slot players, you've got 12 different arrangements corresponding to each of the first-slot players. Your first two could be A and J, or B and J, or C and J, etc. So we'll actually need to multiply the number of options for each slot together with the numbers of options for all other slots, like this:

12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4

That will give us our full number of combinations. So the full number of different possible lineups will come out to whatever huge number that multiplies together to make.

At that point, you'll just need to divide that huge number by the number of innings in a game to find out how many games you can fill, and then again by the number of games in a season to figure out how many seasons.

Hope this helped!