Dtjytyk

I managed to get

$$3(2^{10} -1) - 3(2^{5}-1) - 3(2^{5}-1) $$

or

$$3(2^{10} -1) - 3(2^{5} + 2^{5}-1) $$

However, I am a little confused why the answer isn't

$$3(2^{10} -1) - 3(2^{5}) - 3(2^{5}) $$

Why do we have to have $-1$ in all cases? Wouldn't it be enough to have $-1$ only once so we get rid of the empty case once?

Let's use the Inclusion-Exclusion Principle to solve the problem.

There are $3$ ways to choose the size of the pizza and $2^{10}$ ways to choose which subset of toppings will be placed on the pizza. Hence, if there were no restrictions, we could select a pizza in $3 cdot 2^{10}$ ways.

From these, we must subtract those selections in which no meat toppings or no veggie toppings are selected.

No meat toppings: There are again $3$ ways to choose the size of the pizza. If there are no meat toppings, only the five veggie toppings can be selected. There are $2^5$ ways to choose a subset of the veggie toppings. Thus, there are $3 cdot 2^5$ ways to choose a pizza with no meat toppings.

No veggie toppings: By symmetry, there are $3 cdot 2^5$ ways to choose a pizza with no veggie toppings.

If we subtract the number of pizzas with no meat toppings and the number of pizzas with no veggie toppings from the total, we will have subtracted those pizzas with no meat and no veggie toppings twice. We only want to subtract them once, so we must add them back.

No meat and no veggie toppings: There are only three such pizzas, one of each size.

Total: There are $$3 cdot 2^{10} - 3 cdot 2^5 - 3 cdot 2^5 + 3 = 3(2^{10} - 2^5 - 2^5 + 1)$$
pizzas that have at least one meat topping and at least one veggie topping. Notice that
$$3(2^{10} - 2^5 - 2^5 + 1) = 3(2^{10} - 1) - 3(2^5 - 1) - 3(2^5 - 1)$$
Why should this be the case?

We want pizzas with at least one meat topping and at least one veggie topping. The term $3(2^{10} - 1)$ counts all pizzas with at least one topping. One of the terms $3(2^5 - 1)$ counts all pizzas with at least one topping that contain no meat toppings, and the other one counts all pizzas with at least one topping that contain no veggie toppings. The $2^5 - 1$ terms are necessary since you started by counting all pizzas with at least one topping. If you instead start with all possible pizzas, you arrive at the answer obtained above with the Inclusion-Exclusion Principle.

There are $2^5-1$ ways to choose at least one meat and similarly for veggies, and the answer is $3(2^5-1)^2$.

For one size, refer to the table ($0$ - absent, $1$ - present):
$$begin{array}{c|c|c}
text{Cases}&text{Meat}&text{Veggi}&text{All}&text{All but $00$}\
hline
(1)&0 text{or} 1&0 text{or} 1&00+01+10+11& 2^{10}-1\
(2)&0&0 text{or} 1&00+01& 2^{5}-1\
(3)&0 text{or} 1&0&00+10& 2^{5}-1\
hline
(1)-(2)-(3)&&&&(2^{10}-1)-(2^5-1)-(2^5-1)end{array}$$
For three sizes, it must be multiplied by $3$.