Problem 14, Ch. 1 from Blitzstein and Hwang, Intro to Probability 
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 also allowed, as is getting all of 8). How many possibilities are there for your two pizzas?

My attempt:
Denoting 0:excluding a topping, 1:including a topping, the number of possible combinations of toppings is the same as sampling from the set $\{0,1\}$ with replacement and with ordering, 8 times; there are $2^8=256$ possible toppings combinations.  
Choosing the pizzas is the same as sampling from the set {1,2,3,4} with replacement and with ordering, 2 times; there are $4^2=16$ possible pizza combinations.
The 256 possible toppings need to be assigned to 16 pizza combinations. This amounts to sampling from $\{1,2,3,\ldots,256\}$, 16 times without replacement, but with ordering. Thus, there are $n{\cdot}(n-1){\ldots}(n-k+1)=\displaystyle{256{\cdot}255{\cdot}{\ldots}{\cdot}241}$ possibilities. 
Assumption: The same topping combination is not allowed for both the pizzas.
Is my line of thinking correct, could someone help validate.
 A: I don't see any reason for assuming that the same topping combination is not allowed for the two pizzas, in the absence of any such stipulation in the question.
Ways of choosing a pizza (size) $= 4$
Ways of choosing toppings for the pizza $= 2^8 = 256$
Thus there are $4\cdot256 = 1024$ types of pizzas that can be ordered.
And since the question mentions your two pizzas,
you can order either two different types or same types in $\binom{1024}{2} + 1024$ ways
A: You are right that there are $256$ different toppings possible. Now the following condition makes the enumeration quite a bit trickier.  

The same topping combination is not allowed for both the pizzas.

I am not sure if this condition is part of the problem or a condition that you added. Either way, I am assuming that you only want to count the combinations where the pizzas have different toppings. 
Let's break up our task into counting the number of combinations where we have (1) same sizes of pizza and (2) different sizes of pizza. 
For (1), we have four ways to choose the common size. Now of the $256$ different topping combinations, we choose 2 of them for our pizzas. Note that ordering a small cheese pizza and small supreme pizza is the same as ordering a small supreme pizza and a small cheese pizza. So there are $4 {256 \choose 2}$ different choices here. 
For (2), we have ${4 \choose 2}$ ways to pick the two different sizes. Now we choose one topping choice for the smaller pizza (in $256$ ways) and then a different topping choice for the other pizza (in $255$ ways). Note that ordering a small cheese pizza and a large supreme pizza is different than ordering a small supreme pizza and a large cheese pizza. Thus there are ${4 \choose 2} (256) (255)$ different choices for this case. 
Overall, there are
$$
4 {256 \choose 2} + {4 \choose 2} (256) (255) = 4*256*255
$$
ways to choose the pair of pizzas. 
