It would be incorrect to consider each tomato slice a separate ingredient, since you can't differentiate between them. All other things being equal, a Whopper with just tomato slice #1 is no different from a Whopper with just tomato slice #2: they both just have one tomato slice on them. Instead, if you have an ingredient with a maximum count of k, then there are k + 1 possibilities corresponding to it: 0, 1, 2, ..., k. There are three possible counts of tomato slices on your Whopper: zero slices, one, or two. Similarly, there are five ways you can have pickle slices on your Whopper.
So that adds up to a somewhat less impressive 26 × 3 × 5 = 960 possible Whoppers. If you consider the top and bottom buns distinct, then that brings in a factor of 2, which brings the total up to 1,920 possible Whoppers.
This is only considering the stuff above the horizontal rule in the diary. If you want to find out how many different combinations of extras you can get with a given amount of money, given the prices of all those extras, it's easy to find with a linear homogeneous recurrence relation, but only if you differentiate between different orders of extras (i.e., it would count once for extra pickles on top of extra onions, and once for extra onions on top of extra pickles). Discounting order would require partitions, though, and partition counting is hard.
However, if there's no maximum price, then there are an infinite variety of Whoppers that you can get, since you can order an arbitrary number of pickle slices, right? They're countably infinite, though, and it appears that the set of all Whoppers would be countably infinite as well, since either you count the order of the ingredients, and they're all finite strings on the alphabet of Whopper ingredients, which is finite, or you don't count order, and they're all sub-multisets of the set of ingredients, which is finite.
harder, better, faster, stronger: e2