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 ${\large[}4{\large[}{8\choose8}+{8\choose7}+{8\choose6}+{8\choose5}+{8\choose4}+{8\choose3}+{8\choose2}+{8\choose1}+{8\choose0}{\large]}{\large]}^2$
Best Answer
You are correct if the order you get your pizzas matters. The stuff inside the outer square brackets (which equals $4 \cdot 2^8=1024$) is the number of single pizzas. It seems more likely that getting pizza A and then B is the same as getting B and then A. In that case you have to divide the cases of disparate pizzas by $2$, so there would be $1024$ (ways to get two the same) + $\frac 12\cdot 1024 \cdot 1023$ (ways to get two different).
You can save yourself computing all the $8 \choose n$'s by noticing that you have eight binary choices to make, one for each topping. You can make them in $2^8$ ways, so that gives the sum of all the $8 \choose n$'s