8C3 = 8*7*6/(3*2*1) = 56
You can make 28 = 256 pizzas. Topping 1: You either have it or you don't: two choices. With each choice, for topping 2: You either have it or you don't: two choices. That makes 2*2 or 22 choices. With each choice for the first two, for topping 3: You either have it or you don't: two choices. That makes 2*2*3 or 23 choices. and so on, making 28 choices in all. Note that one choice will comprise no toppings.
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.
three
2*2*2*2 = 16, counting one with no toppings.
220
4 pizzas
8C3 = 8*7*6/(3*2*1) = 56
they can come in many toppings e.g. pepperoni, chicken and sweetcorn, pineapple and ham etc.
14 x 13 = 182
13
You can make 28 = 256 pizzas. Topping 1: You either have it or you don't: two choices. With each choice, for topping 2: You either have it or you don't: two choices. That makes 2*2 or 22 choices. With each choice for the first two, for topping 3: You either have it or you don't: two choices. That makes 2*2*3 or 23 choices. and so on, making 28 choices in all. Note that one choice will comprise no toppings.
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
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.
120 5 x 4 x 3 x 2 x1 = Toppings 20 x 6 x 1 = Toppings 120 x 1 = Toppings 120 = Topping Combinations
You have six different toppings. You want to make a three topping pizza. You don't want to double any topping. This can be expressed as the number of combinations of 6 things taken 3 at a time. The answer is 20. The general case of the number of combinations of N things taken P at a time is N! / (P! * (N - P)!). N = 6. P is 3. Substituting, we get 720 / (6 * (6)), or 720 / 36, or 20.