10
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
8 over 3, which is the same as (8 x 7 x 6) / (1 x 2 x 3).
13
Two . . . . . 38 and 83.
16 i think
32 combinations. 4 of these will have no toppings, or all three toppings, 12 will have one topping and another 12 will have 2 toppings.
it is i love hunter elam
10
well, you can to topping 1&2, topping 2&3, topping 1&3, topping 1, 2 and 3, and you can also do all three toppings. so that's seven different types for one size pizza, and you can have all combinations in four sizes. that makes a total of 28 different pizza combinations.
2*2*2*2 = 16, counting one with no toppings.
three
16 1 combination of all 4 4 combinations of 3 6 combinations of 2 4 combinations of 1 1 combination of 0
18
36
47
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