Any maths buffs?

what about AABB?

and is there a max toppings per pizza limit?
and do doubles count twice against that limit?

Forgot AABB, it counts.
Don't think there is a max toppings per pizza.

why would you want pineapple twice on your pizza ?

because is ace!

Anyway, ignoring the double toppings, the answer is this:

n = total number of toppings available
r = number of toppings on current pizza combination
t = total number of pizza choices

t = Σ(r = 0 to n) nCr

Double toppings add some complexity, as cheese_1 = cheese_2 when used alone (i.e. pizza {pepper,cheese_1} = pizza {pepper,cheese_2}, but neither are equivalent to pizza {pepper,cheese_1,cheese_2})

So for each pizza with r toppings, there are 2^r possible pizzas with double toppings

eg. pizza = {ham,pineapple}
can also be
p_2 = {ham,double pineapple}
p_3 = {double ham, pineapple}
p_4 = {double ham,double pineapple}

So for n toppings, there are:
t = (
(nC0) * 2^0

• (nC2) * 2^2
  
• (nC3) * 2^3
  ...
  
• (nCn) * 2^n
  )
  
  

pizzas.

I think.

I don't think this works when fed back the data....