Palatable Pizzeria Pizzas

First note that all 3 pizzas cannot be the same due to the different toppings.

[Case 0 : 3 pizzas are the same. This is not possible. - Calvin]

Case 1: 2 pizzas are the same.
As all the toppings are different, the only possibility is that two pizzas have no topping and all the toppings are on 1 pizza which contributes one set.

Case 2: All pizzas are distinct.
Assume we number the pizzas 1 to 3. We can place each topping on one of 3 pizzas giving us $3^7$ possible triplets. Of these triplets, 3 of them involve the first case which is when all the toppings are on pizza 1,2 or 3. As the order of the pizzas does not matter, we have a total of $\frac{3^7-3}{3!}=364$ sets.

In total, we have $1+364=365$ sets

There are 8 possibilities for the number of toppings on each pizza. They are: 7, 0, 0 6, 1, 0 5, 1, 1 5, 2, 0 4, 2, 1 4, 3, 0 3, 3, 1 3, 2, 2 Each sequence of three numbers corresponds to the number of toppings on each pizza in non-increasing order. Now we count the number of possible sets of pizzas for each case and sum them up. Case 1: 7, 0, 0 Trivially 1 Case 2: 6, 1, 0 7C6 = 7 Case 3: 5, 1, 1 First we choose 5 out of 7 toppings. The remaining two only has 1 way to split to 2 pizzas. 7C5 = 21 Case 4: 5, 2, 0 Choose 5 out of 7 toppings. The other 2 go onto one pizza. 7C5 = 21. Case 5: 4, 2, 1 Choose 4 out of 7 toppings. Then choose 1 out of 3 remaining toppings to go on one pizza. The remaining 2 toppings will go onto one pizza. 7C4 * 3 = 105 Case 6: 4, 3, 0 Choose 4 out of 7 toppings. The remaining 3 will have to stick together. 7C4 = 35 Case 7: 3, 3, 1 Choose 1 out of 7 toppings. From the remaining 6, choose 3. Note, however, that choosing one group of 3 is the same as choosing the remaining group of 3. Hence our count for the number of possible sets of pizzas is 7 * 6C3 / 2 = 70 Case 8: 3, 2, 2 Choose 3 out of 7. From the remaining 4 choose 2. Again, choosing 2 is the same as choosing the remaining 2. Hence we get 7C3 * 4C2 / 2 = 105. The sum of the number of pizza sets from each case is 365.

We have the following possible cases:
- A first pizza with a type of topping, a second also with a type and a third with 5 types of toppings, having thus $P ^ 5_7 = 42$ sets;
- A first pizza with a type of topping, a second with 2 types and a third with 4 types, having thus $P ^ {2.4} _ 7 = 105$ sets;
- A first pizza with a type of topping, a second with 3 types and a third with 3 types, having thus $P ^ {3.3} _ 7 = 140$ sets;
- A first pizza with 2 kinds of toppings, a second also with 2 types and a third with 3 types of toppings, having thus $P ^ {2,3,3}_7 = 70$ sets;
- A first pizza with no topping, a second also with no topping and a third with 7 kinds of toppings, and thus a set only;
- A first pizza with no topping, a second with a kind of topping and a third with 6 types, having thus $P ^ 6_7 = 7$ sets.

Therefore qualify $42 + 105 + 140 + 70 + 1 + 7 = 365$ sets for the promotion.

If we label the pizzas, then there are $3^7 = 2187$ ways of assigning the toppings to the pizzas in a way that qualifies for the promotion. However, this will count each combination of pizzas multiple times. There are three orders where all the toppings are on a single pizza, and the other two pizzas both have no toppings. That leaves $2184$ orders where no pair of pizzas have the same set of toppings. Each of these orders occurs $6$ times, since there are $3!$ ways of getting that set of pizzas. This means there are $\frac{2184}{6} = 364$ distinct orders where at most one pizza has no toppings, and 1 order where two pizzas have no toppings, for a total of $365$ orders that qualify for the promotion.