/*! 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 5 Evaluate the permutations. $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the permutations. $$ P_{3}^{5} $$

Short Answer

Expert verified
Answer: There are 60 different ways to arrange 5 objects, taken 3 at a time.

Step by step solution

01

Identify the values of n and r

In this exercise, we are given $$n=5$$ and $$r=3$$ from the permutation $$P_3^5$$.
02

Use the permutation formula

We will use the formula for a permutation: $$ P_{n}^{r} = \frac{n!}{(n-r)!} $$ Substituting the values of $$n$$ and $$r$$, the permutation is: $$ P_{3}^{5} = \frac{5!}{(5-3)!} $$
03

Calculate the factorial values

Now, calculate the factorials: $$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$ and $$ (5-3)! = 2! = 2 \times 1 = 2 $$
04

Substitute the factorial values and find the result

Substitute the factorial values into the formula and solve: $$ P_{3}^{5} = \frac{5!}{(5-3)!} = \frac{120}{2} = 60 $$ The permutation $$P_3^5$$ equals 60, which means there are 60 different ways to arrange 5 objects, taken 3 at a time.

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Key Concepts

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

Factorial
Understanding factorials is fundamental in various branches of mathematics, including permutations and combinatorics. A factorial, represented by an exclamation mark (!), is the product of all positive integers less than or equal to a given number. For example, the factorial of 5, written as \(5!\), is calculated as:

\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
It's important to remember that \(0!\) is defined as 1. Factorials grow very quickly with larger numbers, a concept known as 'factorial explosion'. In permutations, factorials determine the total number of ways objects can be arranged. Practicing the calculation of smaller factorials helps students become familiar with the concept and prepares them for tackling larger and more complex problems.
Permutation Formula
Permutation, a key concept in combinatorics, refers to the arrangement of objects in a particular order. The permutation formula is used to calculate the number of possible arrangements when order matters. The generalized form is:

\[P_n^r = \frac{n!}{(n-r)!}\]
where \(n\) is the total number of objects, and \(r\) is the number of objects we want to arrange. When applying the permutation formula, as seen in the example with \(P_3^5\), it's crucial to subtract the base number by the number we are arranging to find the denominator of our fraction. Understanding how factorials play into this formula, as they represent the number of arrangements, is critical for fully grasping the concept. By plugging in the factorial values, one can derive the total number of distinct permutations.
Combinatorics
Combinatorics is a vast area of mathematics focused on the counting, arrangement, and combination of objects. Within it, the study of permutations examines how to arrange objects in a sequence where order is important. The difference between combinations and permutations is that combinations do not consider order, while permutations do. This distinction is essential when solving problems related to probability, game theory, and other mathematical fields.

Understanding the basic principles of combinatorics is beneficial not only in pure mathematics but also in practical applications such as cryptography, network design, and algorithm development. To develop a strong foundation in combinatorics, students should practice problems involving permutations and combinations, always paying close attention to whether the order of arrangement or selection is significant in the context of the problem.

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Most popular questions from this chapter

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. A fair die is tossed twice. What is the probability that the sum of the two dice is \(11 ?\)

Experiment III A sample space consists of five simple events with \(P\left(E_{1}\right)=P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right) .\) Find the probabilities for simple events \(E_{4}\) and \(E_{5}\).

A sample is selected from one of two populations, \(S_{1}\) and \(S_{2},\) with \(P\left(S_{1}\right)=.7\) and \(P\left(S_{2}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}\) or \(S\), has occurred are $$ P\left(A \mid S_{1}\right)=.2 \text { and } P\left(A \mid S_{2}\right)=.3 $$ Use this information to answer the questions in Exercises \(1-3 .\) Use the Law of Total Probability to find \(P(A)\).

Refer to Exercise 33. Suppose that there are six prospective jurors, four men and two women, who might be chosen for the jury. Two jurors are randomly selected from these six to fill the two remaining jury seats. a. List the simple events in the experiment (HINT: There are 15 simple events if you ignore the order of selection of the two jurors.) b. What is the probability that both impaneled jurors are women?

When an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2}\) and \(S_{3}\), can occur, with \(P\left(S_{1}\right)=.2, P\left(S_{2}\right)=.5,\) and \(P\left(S_{3}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}, S_{2}\), or \(S_{3}\) has occurred are $$ P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3 $$ If event A is observed, use this information to find the probabilities in Exercises 4 -6. \(P\left(S_{1} \mid A\right)\)

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