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A pizza establishment offers 12 vegetable toppings and 5 meat toppings. Find the number of different ways you could select a. A meat topping or a vegetable topping b. A meat topping and a vegetable topping

Short Answer

Expert verified
There are 17 ways to select a meat or vegetable topping and 60 ways to select one meat and one vegetable topping.

Step by step solution

01

Calculate Selections for Meat or Vegetable Topping

To calculate the number of ways to choose a meat topping or a vegetable topping, we add the number of meat toppings to the number of vegetable toppings. This is because we want either one or the other, not both.
02

Calculate Selections for Meat and Vegetable Topping

To calculate the number of ways to choose both a meat topping and a vegetable topping, we multiply the number of meat toppings by the number of vegetable toppings. This is because we want one of each category, and every meat topping can be paired with each vegetable topping.
03

Perform the Calculations

For a meat or vegetable topping: There are 12 vegetable toppings + 5 meat toppings = 17 possible choices.For a meat and vegetable topping: We have 12 vegetable toppings * 5 meat toppings = 60 possible combinations of one meat and one vegetable topping.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations and Combinations
Understanding the concepts of permutations and combinations is essential for solving problems involving the arrangement and selection of objects. Simply put, permutations are about arranging items in order, while combinations are about selecting items without regard to order.

For example, if our pizza establishment has 5 types of meat toppings and customers care about the order in which they are added to the pizza, then the number of different permutations of these toppings for a pizza with two types of meat would be calculated using the permutation formula: \( P(n, r) = \frac{n!}{(n-r)!} \), where 'n' is the total number of items to choose from, 'r' is the number of items to choose, and '!' denotes factorial.

However, in most cases, like our pizza problem, order doesn't matter, and that's where combinations come into play. To choose a topping without worrying about the order, we would use the combination formula: \( C(n, r) = \frac{n!}{r!(n-r)!} \). This provides us with a way to calculate the number of possible unique sets of toppings when the order is irrelevant, which is more common in selection problems like this.
Principles of Counting
The principles of counting are fundamental tools in combinatorics that help us enumerate possible outcomes without having to list them all out. These include the addition and multiplication principles, which were applied in our pizza topping problem.

The addition principle states that if we have a choice between one action that can be done in 'm' ways and another that can be done in 'n' ways, and the two actions cannot occur at the same time (they are mutually exclusive), then there are \( m + n \) total ways to choose an action. This principle was used in Step 1 when we added together the meat and vegetable topping options because a customer would choose either a meat or a vegetable topping, but not both at once.

Conversely, the multiplication principle tells us that if one action can be performed in 'm' ways and a second action can be done in 'n' ways after the first action has been completed, then there are a total of \( m \times n \) ways to perform both actions in succession. This was used in Step 2, where we are considering the selections of one meat and one vegetable topping together, allowing every combination of the two categories to come into play.
Probability
Probability is the measure of how likely it is for a particular event to occur. In essence, it quantifies uncertainty and is a fundamental concept not just in mathematics, but in various fields including science, finance, and philosophy.

When applied to simple events, probability can be calculated as the number of favorable outcomes divided by the total number of possible outcomes. In the context of our pizza establishment, if a customer were to randomly select a topping, the probability of choosing a particular type (meat or vegetable) would be determined by the ratio of that type's toppings to the total number of toppings available. This connection between probability and counting principles highlights why understanding permutations and combinations is vital to mastering probability problems.

While probability wasn't directly calculated in our example problem, if we were asked to find the likelihood of picking a specific combination of meat and vegetable toppings, we would use the principles of probability in tandem with previously calculated combinations to determine our answer.

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Most popular questions from this chapter

Car Breakdown Problem: Suppose you plan to drive on a long trip. The probability your car will have a flat tire is \(0.1,\) and the probability it will have engine trouble is \(0.05 .\) What is the probability of each of these events? a. No flat tire b. No engine trouble c. Neither flat tire nor engine trouble d. Both a flat tire and engine trouble e. At least one, either a flat tire or engine trouble

First Girl Problem: Eva and Paul want to have a baby girl. They know that the probability of having a girl on any single birth is 0.5 (A). Let \(x\) be the number of babies they have, and let \(P(x)\) be the probability that the \(x\) th baby is the first girl. Then \(P(1)=0.5 . P(2)\) is the probability that the first baby is not a girl and that the second baby is a girl. Calculate \(P(2), P(3),\) and \(P(4)\) (B). Plot the graph of \(P\). Sketch the graph, showing what happens as \(x\) becomes large. (C). Besides being called a probability distribution, what other special kind of function is this? (D). Show that the sum of the values of \(P(x)\) approaches 1 as \(x\) becomes very large.

Baseball Team Problem 1: Nine people on a baseball team are trying to decide who will play each position. a. In how many different ways could they select a person to be pitcher? b. After someone has been selected as pitcher, in how many different ways could they select someone to be catcher? c. In how many different ways could they select a pitcher and a catcher? d. After the pitcher and catcher have been selected, in how many different ways could they select a first-base player? e. In how many different ways could they select a pitcher, catcher, and first- base player? f. In how many different ways could all nine positions be filled? Surprising?!

Journal Problem: Update your journal with things you have learned since the last entry. In particular, tell how large numbers of outcomes can be calculated, rather than counted, using the concepts of factorials, combinations, and permutations.

A salesperson has 7 customers in Denver and 13 customers in Reno. In how many different ways could she telephone a. A customer in Denver and then a customer in Reno? b. A customer in Denver or a customer in Reno, but not both?

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