Zoom International Math League

Just to make sure we're on the same page for the $120$ using cases (and in case others are reading the discussion):

• With three different toppings we have $\displaystyle \binom{8}{3} = 56$ different pizzas.
• With two different toppings (so one topping double and the other single) we have $8\cdot 7 = 56$ different pizzas ($8$ choices for the double topping, $7$ for the single topping).
• If there's just one (triple) topping: $8$ ways.

This gives $56+56+8 = 120$ different choices for pizzas.

Note when doing these cases, we only care about which toppings are used, so the toppings are not in any particular order. In fact, this means we can use stars and bars (or balls and boxes) here. Think of $3$ identical balls (one for each topping) and then we put those balls into $8$ boxes (labeled for the different toppings). Hence, we have $3$ stars and $8-1 = 7$ bars, with $\displaystyle \binom{10}{3} = \frac{10!}{3!7!} = 120$ total outcomes.

Hope this makes sense! Let us know if you have any other additional questions.