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

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

2

toppings?

\newline

Answer:

Calculate Combinations Formula: We need to calculate the number of combinations of $2$ toppings that can be made from $6$ different toppings. The order in which we select the toppings does not matter, so we will use the combination formula which is given by: $\newline$ $C(n, k) = \frac{n!}{k!(n - k)!}$ $\newline$ where $n$ is the total number of items to choose from, $k$ is the number of items to choose, $n!$ denotes the factorial of $n$ , and $C(n, k)$ denotes the number of combinations.
Identify Values: First, we identify the values of $n$ 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 Formula: Now we apply the values to the combination formula: $\newline$ $C(6, 2) = \frac{6!}{(2!(6 - 2)!)}$ $\newline$ $C(6, 2) = \frac{6!}{(2! \cdot 4!)}$ $\newline$ We know that $6! = 6 \cdot 5 \cdot 4!$ , $2! = 2 \cdot 1$ , and $4!$ cancels out on the numerator and denominator.
Simplify Expression: Simplifying the expression, we get: $\newline$ $C(6, 2) = \frac{6 \times 5}{2 \times 1}$ $\newline$ $C(6, 2) = \frac{30}{2}$ $\newline$ $C(6, 2) = 15$
Final Result: Therefore, there are $15$ different ways to make a pizza with $2$ toppings from a selection of $6$ toppings.

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