possible pizza orders You are ordering two pizzas. A pizza can be small, medium, large, or extra large, with any combination of 8 possible toppings (getting no toppings is allowed, as is gettting all 8). How many possibilities are there for your two pizzas?
Would it be ${\large[}4{\large[}{8\choose8}+{8\choose7}+{8\choose6}+{8\choose5}+{8\choose4}+{8\choose3}+{8\choose2}+{8\choose1}+{8\choose0}{\large]}{\large]}^2$

Let $i$ be the number of toppings you wish to include on your pizza, where $0 \leq i \leq 8$.
Choosing one pizza could be calculated as follows, $$4\sum\limits_{i=0}^8 {8\choose i} = 4\cdot2^8$$
To calculate the number of possibilities, I will count the number of ways in which you can have the same pizza plus the number of ways in which you can have two different pizzas:
Possibilities for choosing the same pizza is equal to the number of different possibilities for one pizza, that is  $$4\sum\limits_{i=0}^8 {8\choose i} = 4\cdot2^8$$
Counting the possibilities for two different pizzas as follows, $$\frac{\Bigg[4\sum\limits_{i=0}^8 {8\choose i}\Bigg]\cdot \Bigg[\Bigg(4\sum\limits_{i=0}^8 {8\choose i}\Bigg)-1\Bigg]}{2}$$
$$= \frac{(4\cdot 2^8)\cdot(4\cdot 2^8 -1)}{2}$$
You're dividing by two since the order in which you pick the pizzas are not relevant.
In total, that gives you $$(4\cdot 2^8) + \frac{(4\cdot 2^8)\cdot(4\cdot 2^8 -1)}{2} = 524800$$ possibilities for two pizzas.
