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Chapter 1: Problem 38

In how many ways may \(n\) people sit around a round table? (Assume that when people are sitting around a round table, all that really matters is who is to each person's right. For example, if we can get one arrangement of people around the table from another by having everyone get up and move to the right one place and sit back down, we get an equivalent arrangement of people. Notice that you can get a list from a seating arrangement by marking a place at the table, and then listing the people at the table, starting at that place and moving around to the right.) There are at least two different ways of doing this problem. Try to find them both. (h)

Short Answer

Expert verified
(n-1)!

Step by step solution

01

Fix One Person

Since the seats are arranged in a circle, we can fix one person in one seat to remove the effect of the circular permutation. Suppose we fix one person in a particular seat.
02

Arrange Remaining n-1 Persons

After fixing one person, we need to arrange the remaining \(n-1\) people. The number of ways to do this is \( (n-1)! \).
03

Generalize the Result

Hence, the number of ways to arrange \(n\) people around a round table is \( (n-1)! \)

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Key Concepts

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

circular permutations
When it comes to arranging people or objects in a circle, we deal with a special type of permutation known as circular permutations. Unlike linear permutations where the order matters from left to right, in circular permutations, what matters is the relative positioning of the items. Essentially, if you can rotate one arrangement to get another, those two arrangements are considered the same.

To simplify the problem of counting circular permutations, we can fix one person in place and then arrange the others around them. This way, we've effectively converted a circular permutation into a linear permutation of the remaining items.
factorial
A factorial, represented by the symbol \, is a mathematical function that multiplies a number by all the positive integers less than itself. For example, \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \). Factorials are key in permutation and combination problems because they represent the number of ways to arrange a set of items.
In our problem, after fixing one person to break the circle, we are left with \(n-1\) people to arrange. The number of ways to arrange these \(n-1\) people is represented by \((n-1)!\). Thus, the formula for the number of ways \(n\) people can sit around a round table is \((n-1)!\).
Remember that the key to understanding factorials is recognizing they represent a sequence of multiplications that describe arrangements.
combinatorics problem-solving strategies
Combinatorics, the branch of mathematics dealing with combinations and permutations, provides powerful strategies for solving problems involving arrangements. One common strategy is **fixing** an element to simplify circular permutations. By fixing one person, the problem converts from a circular to a linear permutation.

Another strategy is **breaking down complex problems** into smaller, manageable steps, just like the exercise does. First, we fix one person; then we consider the arrangement of the remaining \(n-1\) people.
Finally, always verify that different methods yield the same result. This reinforces your understanding and ensures accuracy. In this problem, both fixing one person and recognizing rotational symmetry should lead to the same result, \((n-1)!\).

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Most popular questions from this chapter

As we noted in Problem \(29,\) the first question in Problem 8 asked us for the number of three-element subsets of a twelve-element set. We were able to use the Pascal Equation to get a numerical answer to that question. Had we had twenty or thirty flavors of ice cream to choose from, using the Pascal Equation to get our answer would have entailed a good bit more work. We have seen how the general product principle gives us an answer to Problem 6 . Thus we might think that the number of ways to choose a three element set from 12 elements is the number of ways to choose the first element times the number of ways to choose the second element times the number of ways to choose the third element, which is \(12 \cdot 11 \cdot 10=1320\). However, our result in Problem 29 shows that this is wrong. (a) What is it that is different between the number of ways to stack ice cream in a triple decker cone with three different flavors of ice cream and the number of ways to simply choose three different flavors of ice cream? (b) In particular, how many different triple decker cones use the same three flavors? (Of course any three distinct flavors could substitute for vanilla, chocolate and strawberry without changing the answer.) (c) Using your answer from part b, compute the number of ways to choose three different flavors of ice cream (out of twelve flavors) from the number of ways to choose a triple decker cone with three different flavors (out of twelve flavors).

One of the schools sending its team to the tournament has to send its players from some distance, and so it is making sandwiches for team members to eat along the way. There are three choices for the kind of bread and five choices for the kind of filling. How many different kinds of sandwiches are available? (b)

A roller coaster car has \(n\) rows of seats, each of which has room for two people. If \(n\) men and \(n\) women get into the car with a man and a woman in each row, in how many ways may they choose their seats?

In a part of a city, all streets run either north-south or east-west, and there are no dead ends. Suppose we are standing on a street corner. In how many ways may we walk to a corner that is four blocks north and six blocks east, using as few blocks as possible? (h)

Two sets are said to be disjoint if they have no elements in common. For example, \(\\{1,3,12\\}\) and \(\\{6,4,8,2\\}\) are disjoint, but \(\\{1,3,12\\}\) and \(\\{3,5,7\\}\) are not. Three or more sets are said to be mutually disjoint if no two of them have any elements in common. What can you say about the size of the union of a finite number of finite (mutually) disjoint sets? Does this have anything to do with any of the previous problems?
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