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Exploring permutations. The formula for the number of ways to arrange \(n\) objects is \(n !=n \times(n-\) 1) \(\times \cdots \times 2 \times 1\). This exercise walks you through the derivation of this formula for a couple of special cases. A small company has five employees: Anna, Ben, Carl, Damian, and Eddy. There are five parking spots in a row at the company, none of which are assigned, and each day the employees pull into a random parking spot. That is, all possible orderings of the cars in the row of spots are equally likely. (a) On a given day, what is the probability that the employees park in alphabetical order? (b) If the alphabetical order has an equal chance of occurring relative to all other possible orderings, how many ways must there be to arrange the five cars? (c) Now consider a sample of 8 employees instead. How many possible ways are there to order these 8 employees' cars?

Short Answer

Expert verified
(a) Probability is \( \frac{1}{120} \). (b) There are 120 ways to arrange 5 cars. (c) There are 40,320 ways to arrange 8 cars.

Step by step solution

01

Understand the Formula

The number of ways to arrange \( n \) objects is given by the factorial of \( n \), written as \( n! \). This means: \( n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1 \).
02

Calculate the Probability of Alphabetical Order

To determine the probability that the employees park in alphabetical order, first note that there is only 1 way they can park alphabetically (Anna, Ben, Carl, Damian, Eddy). The probability is then 1 out of the total number of arrangements, which is 5!.
03

Compute Total Arrangements for 5 Employees

The total number of ways to arrange 5 employees is given by calculating the factorial of 5: \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). There are 120 different ways for the five employees' cars to be parked.
04

Probability of Alphabetical Order

Since there is only 1 way for the cars to be parked alphabetically out of 120 possible arrangements, the probability is \( \frac{1}{120} \).
05

Calculate Arrangements for 8 Employees

For a group of 8 employees, we use the factorial of 8 to find the total number of arrangements: \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320 \). Thus, there are 40,320 ways to arrange the cars of 8 employees.

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Key Concepts

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

Factorials
Factorials are a mathematical concept used to determine the number of ways to arrange a set of objects.
The factorial of a non-negative integer, say, "n," is expressed as "n!" and calculated by multiplying all whole numbers from "n" down to 1.
  • For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.
  • Similarly, 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320.
Factorials are used to determine possible arrangements, making them a fundamental part of permutation and combination calculations.
In contexts like the parking problem, the factorial tells us how many different sequences can be formed from a set of distinct objects.
Probability
Probability is a measure of the likelihood of an event occurring. In our scenario, we want to explore the probability of employees parking in alphabetical order.
In probability terms, this involves calculating the favorable outcomes versus the total possible outcomes.Imagine for the five employees:
  • Only one way exists to arrange them alphabetically (Anna, Ben, Carl, Damian, Eddy).
  • The total number of arrangements is 5! = 120.
Thus, the probability of this specific arrangement happening is 1 out of 120, or \( \frac{1}{120} \).
Probability can express how rare or common particular arrangements can be, highlighting the significance of each potential outcome.
Arrangement
Arrangement refers to the specific order in which objects or people can be organized. Understanding arrangements is vital when solving permutation-related problems.
For example, in our parking scenario, each unique order of cars represents a different arrangement. When dealing with arrangements:
  • You consider the sequence as each car parks in a different spot.
  • Each different sequence counts as a unique arrangement.
Arrangements are essential in predicting likelihoods and probabilities and are a direct application of permutation principles. They allow us to explore all the possible ways items can be uniquely positioned.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects.
It's the broader umbrella under which concepts like permutations, factorials, and arrangements fall. Through combinatorial methods, we analyze how objects can be arranged and combined:
  • Permutation is an arrangement of objects where order matters.
  • Combination considers sets where order does not matter.
In the parking lot example, we use combinatorics to calculate the number of possible sequences of cars, using factorials to solve permutation problems.
Combinatorics provides the tools and methods needed to navigate and solve complex arrangement challenges, making it indispensable in fields like probability and statistics.

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Most popular questions from this chapter

Chicken pox, Part I. The National Vaccine Information Center estimates that \(90 \%\) of Americans have had chickenpox by the time they reach adulthood. \({ }^{32}\) (a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distribution appropriate for calculating the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood? Explain. (b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpox during childhood. (c) What is the probability that exactly 3 out of a new sample of 100 American adults have not had chickenpox in their childhood? (d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox? (e) What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox?

Game of dreidel. A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting (a) at least one nun? (b) exactly 2 nuns? (c) exactly 1 hei? (d) at most 2 gimels?

Sickle cell anemia. Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, "sickle" shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a \(25 \%\) chance of having the disease, \(50 \%\) chance of being a carrier, and \(25 \%\) chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that (a) two will have the disease? (b) none will have the disease? (c) at least one will neither have the disease nor be a carrier? (d) the first child with the disease will the be \(3^{r d}\) child?

Chickenpox, Part II. We learned in Exercise 4.18 that about \(90 \%\) of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults. (a) How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation? (b) Would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? How does this probability relate to your answer to part (b)?

Male children. While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51 . Suppose a couple plans to have 3 kids. (a) Use the binomial model to calculate the probability that two of them will be boys. (b) Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate the same probability from part (a) but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match. (c) If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, briefly describe why the approach from part (b) would be more tedious than the approach from part (a).

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