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A pizza shop has available toppings of pepperoni, bacon, anchovies, onions, sausage, and peppers. How many different ways can a pizza be made with 3 toppings?
Answer:

A pizza shop has available toppings of pepperoni, bacon, anchovies, onions, sausage, and peppers. How many different ways can a pizza be made with $3$ toppings? $\newline$ Answer:

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Experimental probability

Full solution

Q. A pizza shop has available toppings of pepperoni, bacon, anchovies, onions, sausage, and peppers. How many different ways can a pizza be made with

3

toppings?

\newline

Answer:

Given Toppings: We are given $6$ different toppings to choose from and we want to know the number of different ways to make a pizza with exactly $3$ toppings. This is a combination problem because the order in which we select the toppings does not matter. We can use the combination formula: $\newline$ $C(n, k) = \frac{n!}{k!(n - k)!}$ $\newline$ where $n$ is the total number of items to choose from, $k$ is the number of items to choose, $!$ denotes factorial, and $C(n, k)$ is the number of combinations. $\newline$ In this case, $n = 6$ and $k = 3$ .
Combination Formula: First, we calculate the factorial of $n$ , which is $6!$ . This is equal to $6 \times 5 \times 4 \times 3 \times 2 \times 1$ .
Calculate Factorial: Next, we calculate the factorial of $k$ , which is $3!$ . This is equal to $3 \times 2 \times 1$ .
Calculate Factorial: We also need to calculate the factorial of $n - k$ , which is $6 - 3 = 3$ , so we need to calculate $3!$ again. As we already calculated $3!$ in the previous step, we know it is $3 \times 2 \times 1$ .
Calculate Combination: Now we can plug these values into the combination formula: $\newline$ $C(6, 3) = \frac{6!}{3! * (6 - 3)!}$ $\newline$ $= \frac{(6 * 5 * 4 * 3 * 2 * 1)}{((3 * 2 * 1) * (3 * 2 * 1))}$ $\newline$ $= \frac{(6 * 5 * 4)}{(3 * 2 * 1)}$ $\newline$ $= \frac{120}{6}$ $\newline$ $= 20$ $\newline$ So there are $20$ different ways to make a pizza with $3$ toppings from a selection of $6$ toppings.

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