4 pizzas
25
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if you have 3 different toppings then 12*11*10 = 1320
220
13
4 pizzas
25
There are 7C5 = 7*6/(2*1) = 21 pizzas.
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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
if you have 3 different toppings then 12*11*10 = 1320
as many as you can dream of. each major chain has average of 10 different pizzas. custom ordered pizzas with your choice of toppings. store bought frozen pizzas in endless variety. gourmet pizza shops with their own menus.etc.there are lots of diffrent kinds of pizza
220
To find the number of different 3-topping pizzas that can be made from 21 toppings, we first calculate the combinations of toppings. The number of ways to choose 3 toppings from 21 is given by the combination formula (C(n, k) = \frac{n!}{k!(n-k)!}). Thus, (C(21, 3) = \frac{21!}{3!(21-3)!} = 1330). Since there are 3 different crusts, the total number of 3-topping pizzas is (1330 \times 3 = 3990).
8C3 = 8*7*6/(3*2*1) = 56
three