25
220
16 i think
6666
14 x 13 = 182
13
There are 7C5 = 7*6/(2*1) = 21 pizzas.
If you must use all 5 with no repetition, you can make only one pizza. 5C5, the last entry on the 5 row of Pascal's triangle. If you can choose as many toppings as you want, all the way down to none (cheese pizza), then you have 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 32. Another way to think about it is no toppings would allow one pizza (cheese), one topping would allow two pizzas (cheese, pepperoni), two toppings would allow four pizzas, three toppings would allow eight pizzas, four toppings would allow sixteen, creating an exponential pattern. p = 2 ^ t. So, 10 toppings would permit 1024 different combinations
25
220
8C3 = 8*7*6/(3*2*1) = 56
16 i think
2*2*2*2 = 16, counting one with no toppings.
6666
they can come in many toppings e.g. pepperoni, chicken and sweetcorn, pineapple and ham etc.
14 x 13 = 182
if you have 3 different toppings then 12*11*10 = 1320