/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 SEATING In how many ways can fou... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

SEATING In how many ways can four married couples attending a concert be seated in a row of eight seats if a. There are no restrictions? b. Each married couple is seated together? c. The members of each sex are seated together?

Short Answer

Expert verified
In summary, for the given seating conditions: a. Without any restrictions, there are 40,320 ways to seat the four married couples. b. Ensuring each married couple is seated together, there are 384 ways. c. Seating the members of each sex together, there are 1,152 ways.

Step by step solution

01

Part (a): No restrictions

To find the total number of ways to seat all eight people with no restrictions, we will use the formula for permutations: \(n! = n(n-1)(n-2)...(3)(2)(1)\) Here, since there are 8 people: \(n=8\) So, the total number of ways to seat the 8 people is: \(8! = 8(7)(6)(5)(4)(3)(2)(1) = 40,320\) Thus, there are 40,320 ways to seat the four married couples with no restrictions.
02

Part (b): Each married couple is seated together

First, we'll treat each married couple as a single unit. Since there are 4 couples, there are 4! ways to arrange these couple units: \(4! = 4(3)(2)(1) = 24\) Now, within each couple unit, there are 2 people who can be seated in 2! ways: \(2! = 2(1) = 2\) Since there are 4 couple units, we need to raise 2! to the power of the number of couple units: \( (2!)^4 = 2^4 = 16 \) Multiplying the number of ways to arrange the couple units by the number of ways to arrange the individuals within the units, we get: \(24\times16 = 384\) Thus, there are 384 ways to seat the couples together.
03

Part (c): The members of each sex are seated together

First, we'll arrange the 4 men. There are 4! ways to arrange them: \(4! = 4(3)(2)(1) = 24\) Similarly, there are 4! ways to arrange the 4 women: \(4! = 4(3)(2)(1) = 24\) Since the men and women can be seated together in two different orders (either all the men first followed by all the women, or all the women first followed by all the men), we need to consider this as well. There are 2! ways to determine the order in which the sexes will be seated: \(2! = 2(1) = 2\) To find the total number of ways, we multiply the number of ways of arranging men, women, and their order: \(24\times24\times2 = 1,152\) Thus, there are 1,152 ways to seat the members of each sex together.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Factorial Notation
Factorial notation is fundamental to the study of permutations and combinations. It is denoted by an exclamation point (!) and represents the product of all positive integers up to a given number. For instance, the factorial of 5, written as \(5!\), is calculated as \(5 \times 4 \times 3 \times 2 \times 1 = 120\).

Using factorial notation simplifies the expression and calculation of large numbers involved in permutations, as seen in the earlier problem where the number of ways to arrange eight people is \(8! = 40,320\). Understanding how to compute factorial values is critical for solving combinatorial problems efficiently.

Furthermore, factorial notation is not just limited to positive integers. We can extend the concept to include zero, where \(0!\) is defined to be 1. This particular convention is handy when dealing with permutations and combinations because it simplifies expressions and avoids the need for separate cases when no items are present for arrangement.
Arrangements of Objects
The arrangements of objects, in a specific order, is a problem frequently addressed by permutations. Permutations are about finding the number of ways to order a set of items. When the order matters, the term 'permutation' is used.

For example, when seating people in a theater, we can use permutations to calculate the number of distinct seating arrangements possible. If there are no restrictions, we simply use the factorial of the number of objects. The unrestricted seating arrangement problem for eight people, as in the exercise, is a straightforward permutation problem which uses \(8!\).

However, restrictions can alter the calculation significantly. For instance, if each married couple must sit together, each couple is treated as a single unit, reducing our permutation calculation from eight individuals to four units. Each unit (couple) itself has 2! ways to arrange within the unit, and it's important to calculate for such permutations within these constraints.
Combinatorial Mathematics
Combinatorial mathematics is the field that studies the counting, arrangement, and combination of objects. While permutations deal with the order of arrangements, combinations are concerned with the selection of items where the order is not important.

Without combinations, we'd be left counting the possibilities manually, which is impractical for large sets. In combinations, we utilize binomial coefficients represented by \(C(n, k)\) or \(_nC_k\), which denote the number of ways to choose k items from a set of n items without regard to order.

In our exercise, while the focus was on permutations due to the nature of seating arrangements, combinatorial principles are still at play when considering restrictions such as couples sitting together or members of each sex being seated in their own group. The understanding of combinatorial mathematics is essential to solve these cases where groups within the broader set must be selected or arranged.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fifty raffle tickets are numbered 1 through 50 , and one of them is drawn at random. What is the probability that the number is a multiple of 5 or 7 ? Consider the following "solution": Since 10 tickets bear numbers that are multiples of 5 and since 7 tickets bear numbers that are multiples of 7 , we conclude that the required probability is $$ \frac{10}{50}+\frac{7}{50}=\frac{17}{50} $$ What is wrong with this argument? What is the correct answer?

CoURSE ENROLLMENTS Among 500 freshmen pursuing a business degree at a university, 320 are enrolled in an economics course, 225 are enrolled in a mathematics course, and 140 are enrolled in both an economics and a mathematics course. What is the probability that a freshman selected at random from this group is enrolled in a. An economics and/or a mathematics course? b. Exactly one of these two courses? c. Neither an economics course nor a mathematics course?

A survey in which people were asked how they were planning to prepare their taxes in 2007 revealed the following: $$ \begin{array}{lc} \hline \begin{array}{l} \text { Method of } \\ \text { Preparation } \end{array} & \text { Percent } \\ \hline \text { Computer software } & 33.9 \\ \hline \text { Accountant } & 23.6 \\ \hline \text { Tax preparation service } & 17.4 \\ \hline \text { Spouse, friend, or other } & \\ \text { relative will prepare } & 10.8 \\ \hline \text { By hand } & 14.3 \\ \hline \end{array} $$ What is the probability that a randomly chosen participant in the survey a. Was planning to use an accountant or a tax preparation service to prepare his taxes? b. Was not planning to use computer software to prepare his taxes and was not planning to do his taxes by hand?

An experiment consists of selecting a card at random from a 52-card deck. Refer to this experiment and find the probability of the event. A face card (i.e., a jack, queen, or king) is drawn.

Human blood is classified by the presence or absence of three main antigens (A, B, and Rh). When a blood specimen is typed, the presence of the \(\mathrm{A}\) and/or \(\mathrm{B}\) antigen is indicated by listing the letter \(A\) and/or the letter \(B\). If neither the A nor B antigen is present, the letter \(\mathrm{O}\) is used. The presence or absence of the \(\mathrm{Rh}\) antigen is indicated by the symbols \(+\) or \(-\), respectively. Thus, if a blood specimen is classified as \(\mathrm{AB}^{+}\), it contains the \(\mathrm{A}\) and the \(\mathrm{B}\) antigens as well as the \(\mathrm{Rh}\) antigen. Similarly, \(\mathrm{O}^{-}\) blood contains none of the three antigens. Using this information, determine the sample space corresponding to the different blood groups.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.