/*! 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 27 List all the permutations of fiv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

List all the permutations of five objects \(a, b, c, d,\) and \(e\) taken two at a time without repetition. What is \(_{5} P_{2} ?\)

Short Answer

Expert verified
_{5} P_{2} = 20

Step by step solution

01

Understand the problem

The task requires listing all possible permutations of five objects taken two at a time without repetition, and calculating the permutation value \(_{5} P_{2}\).
02

List permutations

To find the permutations of two objects taken from five without repetition, list all unique pairs:\ \((a, b), (a, c), (a, d), (a, e), (b, a), (b, c), (b, d), (b, e), (c, a), (c, b), (c, d), (c, e), (d, a), (d, b), (d, c), (d, e), (e, a), (e, b), (e, c), (e, d)\). \So, the permutations are: ab, ac, ad, ae, ba, bc, bd, be, ca, cb, cd, ce, da, db, dc, de, ea, eb, ec, ed.
03

Count the total number of permutations

Count the unique permutations listed: there are 20 pairs listed.
04

Apply the permutation formula

The formula for permutations without repetition is \(_{n} P_{r} = \frac{n!}{(n-r)!} \) where n is the total number of objects, and r is the number of objects taken at a time. Here, n = 5 and r = 2. Therefore, \(_{5} P_{2} = \frac{5!}{(5-2)!} = \frac{5!}{3!}\).
05

Calculate the result

Calculate the factorial: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\) and \(3! = 3 \times 2 \times 1 = 6\). Then, \(_{5} P_{2} = \frac{120}{6} = 20\).

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.

Permutation Formula
Permutations are arrangements of objects where the order matters. When we need to find the number of possible permutations of a set of objects, we can use the **Permutation Formula**. For permutations without repetition, the formula is: \[ _{n} P_{r} = \frac{n!}{(n-r)!} \] Here, \( n \) is the total number of objects, and \( r \) is the number of objects we want to choose and arrange.

For example, in our exercise:
  • Total objects \( (n) = 5 \)
  • Number of objects taken at a time \( (r) = 2 \)

Using the permutation formula: \[ _{5} P_{2} = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{120}{6} = 20 \] This tells us there are 20 permutations of five objects taken two at a time without repetition.

The formula helps in quickly finding the number of permutations without needing to list and count them manually.
Factorial
The concept of **factorial**, denoted by \( n! \), is key to understanding permutations. Factorials are used to count the total number of ways to arrange \( n \) objects.

Mathematically, a factorial of a positive integer \( n \) is the product of all positive integers less than or equal to \( n \):
  • \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \)
In our exercise, we calculated:
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \(3! = 3 \times 2 \times 1 = 6 \)


So, when we use the permutation formula, we divide the factorial of the total number of objects by the factorial of the difference between the total objects and the objects taken at a time. This simplifies the process of finding how many unique arrangements can be formed.
Combinatorics
The study of counting, arrangement, and combination of objects is known as **combinatorics**. It's a fundamental area of mathematics that deals with problems of selection and arrangement.

There are several topics within combinatorics, but two of the main ones are:
  • **Permutations**: Arrangements where order matters
  • **Combinations**: Selections where order does not matter
In this exercise, we focused on permutations, specifically:

  • Counting the number of ways to pick and arrange two objects out of five
  • Using the permutation formula to find precise solutions


Combinatorics helps in solving practical problems where we need to determine the number of possible outcomes, such as scheduling, decision-making, and optimization. Understanding these concepts provides a foundation for more advanced studies in probability and statistics.

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

In the U.S. Senate, there are 21 members on the Committee on Banking, Housing, and Urban Affairs. Nine of these 21 members are selected to be on the Subcommittee on Economic Policy. How many different committee structures are possible for this subcommittee?

Among Americans who consider themselves auto racing fans, \(59 \%\) identify NASCAR stock cars as their favorite type of racing. Suppose four auto racing fans are randomly selected. Source: ESPN / TNS Sports, reported in USA Today (a) What is the probability that all four will identify NASCAR stock cars as their favorite type of racing? (b) What is the probability that at least one will not identify NASCAR stock cars as his or her favorite type of racing? (c) What is the probability that none will identify NASCAR stock cars as his or her favorite type of racing? (d) What is the probability that at least one will identify NASCAR stock cars as his or her favorite type of racing?

In \(2004,17 \%\) of Florida's population was 65 and over (Source: U.S. Census Bureau). What is the probability that a randomly selected Floridian is 65 or older?

Suppose that \(E\) and \(F\) are two events and that \(P(E \text { and } F)=0.6\) and \(P(E)=0.8 .\) What is \(P(F | E) ?\)

Determine the probability that at least 2 people in a room of 10 people share the same birthday, ignoring leap years and assuming each birthday is equally likely by answering the following questions: (a) Compute the probability that 10 people have different birthdays. (Hint: The first person's birthday can occur 365 ways; the second person's birthday can occur 364 ways, because he or she cannot have the same birthday as the first person; the third person's birthday can occur 363 ways, because he or she cannot have the same birthday as the first or second person; and so on.) (b) The complement of "10 people have different birthdays" is "at least 2 share a birthday." Use this information to compute the probability that at least 2 people out of 10 share the same birthday.

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.