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A pizza shop has available toppings of mushrooms, peppers, onions, olives, anchovies, and sausage. How many different ways can a pizza be made with 2 toppings?
Answer:

A pizza shop has available toppings of mushrooms, peppers, onions, olives, anchovies, and sausage. How many different ways can a pizza be made with $2$ toppings? $\newline$ Answer:

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Experimental probability

Full solution

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

2

toppings?

\newline

Answer:

Given Toppings: We are given $6$ different toppings to choose from: mushrooms, peppers, onions, olives, anchovies, and sausage. We want to find out how many different combinations of $2$ toppings can be made from these $6$ options. The order in which we select the toppings does not matter (i.e., mushrooms and peppers is the same as peppers and mushrooms), so we will 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, and $k$ is the number of items to choose.
Identify $n$ and $k$ : First, we identify $n$ and $k$ for our problem. We have $n = 6$ toppings to choose from, and we want to choose $k = 2$ toppings for the pizza.
Apply Combination Formula: Now we apply the combination formula: $\newline$ $C(6, 2) = \frac{6!}{2!(6 - 2)!}$ $\newline$ $= \frac{6!}{2! \times 4!}$ $\newline$ $= \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!}$ $\newline$ Since $4!$ cancels out on the numerator and denominator, we simplify the expression.
Simplify Expression: After canceling out $4!$ , we are left with: $\newline$ $C(6, 2) = \frac{6 \times 5}{2 \times 1}$ $\newline$ $= \frac{30}{2}$ $\newline$ $= 15$ $\newline$ So, there are $15$ different ways to make a pizza with $2$ toppings from the given selection.

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