Problem 19
Question
In how many ways can Al, Bob, Carol, Dawn, and Ed be seated in a row of five chairs if \(A l\) is to be seated in the middle chair? 24
Step-by-Step Solution
Verified Answer
There are 24 ways to seat them with Al in the middle.
1Step 1: Examine the Problem Statement
We need to find the number of ways to arrange five people (Al, Bob, Carol, Dawn, and Ed) in a row when Al is fixed in the middle chair. This means that there are four remaining chairs for the rest of the individuals.
2Step 2: Fix Al's Position
Since Al must be in the middle chair, he is already placed, reducing the problem to arranging the remaining four individuals (Bob, Carol, Dawn, and Ed) in the four chairs around him.
3Step 3: Calculate Permutations for Remaining Seats
The four remaining people can be seated in the remaining four seats in different ways. This is a permutation problem for four distinct individuals, which can be calculated as the factorial of 4.
4Step 4: Perform Calculation
Compute the factorial of 4, which is denoted as 4!. This is calculated as follows: \[4! = 4 \times 3 \times 2 \times 1 = 24\]
5Step 5: Conclude the Solution
Since there are 24 ways to arrange the remaining four people, the total number of ways to arrange everyone with Al in the middle is 24.
Key Concepts
FactorialCombinatoricsArrangement Problems
Factorial
The concept of a factorial is crucial when dealing with permutations and arrangement problems. Factorial, denoted by an exclamation mark \(n!\), represents the product of all positive integers up to a given number, \(n\). For example, the factorial of 4, written as \(4!\), is calculated by multiplying 4 by every positive integer less than it: \[4! = 4 \times 3 \times 2 \times 1 = 24\]
Factorials are a fundamental building block in combinatorics and are used to determine the number of ways to arrange a set of objects. They grow rapidly with increasing \(n\). For instance, \(5!\) equals \(120\), and \(6!\) equals \(720\). Understanding how to calculate factorials is essential for solving permutation-based problems effectively.
Factorials are a fundamental building block in combinatorics and are used to determine the number of ways to arrange a set of objects. They grow rapidly with increasing \(n\). For instance, \(5!\) equals \(120\), and \(6!\) equals \(720\). Understanding how to calculate factorials is essential for solving permutation-based problems effectively.
Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arrangement, and combination of objects. It plays a significant role in solving problems related to permutations and arrangements, like the one involving Al, Bob, Carol, Dawn, and Ed seating arrangement.
Combinatorics mainly handles two kinds of problems:
Combinatorics mainly handles two kinds of problems:
- Permutations: These deal with the arrangement of objects in a specific order. For instance, if you're arranging four people in a row, you use factorials to find the number of permutations.
- Combinations: These involve selecting items from a group, where the order does not matter. While combinations are not directly used in our seating problem, understanding both concepts enriches your understanding of grouping problems.
Arrangement Problems
Arrangement problems, also known as permutation problems, involve finding the number of ways to order or arrange a set of objects. In our exercise, arranging the 5 individuals in chairs with Al fixed in the middle highlights a common type of arrangement problem.
Here are some essential aspects:
Here are some essential aspects:
- Fixed Positions: Sometimes, like with Al, certain positions are predetermined, which simplifies the problem by reducing the number of positions that need arranging.
- Permutations: These are calculated using factorial to determine the arrangements for remaining objects, offering a practical way to solve problems efficiently.
Other exercises in this chapter
Problem 19
In how many ways can \(\mathrm{Al}, \mathrm{Bob}\), Carol, Dawn, and Ed be seated in a row of five chairs if \(\mathrm{Al}\) is to be seated in the middle chair
View solution Problem 19
Solve each problem.One card is randomly drawn from a deck of 52 playing cards. What is the probability that it is not an ace? \(\frac{12}{13}\)
View solution Problem 20
Four girls and three boys are to be randomly seated in a row of seven seats. Find the probability that the girls and boys will be seated in alternating seats. \
View solution Problem 20
Six coins are tossed. Find the probability of getting at least two heads. \(\frac{57}{64}\)
View solution