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Chapter 7: Problem 8

Evaluate the given expression. $$ P(5,3) $$

Short Answer

Expert verified
There are 60 ways to arrange 3 objects out of a set of 5 objects using the permutation formula \(P(5, 3) = \frac{5!}{(5-3)!} = \frac{120}{2} = 60\).

Step by step solution

01

Determine the values of n and r

We are given the permutation expression: $$ P(5, 3) $$ This means that we have n = 5 objects and r = 3 objects to arrange.
02

Use the Permutation formula

We will use the formula for Permutations to solve the expression: $$ P(n, r) = \frac{n!}{(n-r)!} $$ For our case, n = 5 and r = 3. We will plug in these values to calculate the Permutation: $$ P(5, 3) = \frac{5!}{(5-3)!} $$
03

Evaluate factorials

Now, we will evaluate the factorials in the expression: 5! means 5 factorial, which is the product of all the positive integers up to 5: $$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$ (5-3)! means (2) factorial, which is the product of all the positive integers up to 2: $$ (5-3)! = 2! = 2 \times 1 = 2 $$ Now we can update the expression: $$ P(5, 3) = \frac{120}{2} $$
04

Calculate the final answer

Finally, we will divide 120 by 2 to find the Permutation: $$ P(5, 3) = 120 \div 2 = 60 $$ Thus, there are 60 ways to arrange 3 objects out of a set of 5 objects.

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Key Concepts

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

Factorials
Factorials are an essential concept in mathematics, especially when dealing with permutations and combinations. A factorial, denoted by an exclamation point (!), represents the product of all positive integers up to a given number. For example, the factorial of 5, written as \(5!\), is calculated by multiplying all whole numbers from 1 through 5: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

Factorials are important because they help us determine the number of possible arrangements or sequences that can be made from a set of objects. This is particularly useful in areas like probability, statistics, and various fields of combinatorial mathematics.
  • Starting point: To calculate \(n!\), begin at the number \(n\).
  • Step downward: Multiply by each successive integer less than \(n\) until you reach 1.
  • Zero factorial: By definition, \(0! = 1\), as there is exactly one way to arrange nothing.
Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arrangement, and combination of objects. It provides tools for counting the number of ways things can be chosen, arranged, or grouped under certain conditions.

A key concept within combinatorics is permutations, which involve arranging a subset of objects from a larger set, where the order matters. The formula for finding permutations is \(P(n, r) = \frac{n!}{(n-r)!}\), where \(n\) is the total number of objects to choose from, and \(r\) is the number of objects to arrange. In our exercise, we used this to find the number of ways to arrange 3 objects from a set of 5.
  • Permutations: Arrangement of objects where the order is important.
  • Combinations: Grouping of objects where the order is not important.
Understanding combinatorics is crucial for solving problems in probability, optimizing decisions, and performing statistical calculations.
Mathematics Education
Teaching concepts like permutations and factorials in mathematics education is vital for building a strong foundation in problem solving and analytical thinking. These concepts are not only essential in mathematics but also in related fields such as computer science and physics.

Incorporating hands-on activities and visual aids can significantly enhance the learning experience. For instance, using physical objects to model permutations or combinations can help students visualize how arrangements work, making the instruction more engaging and accessible.
  • Interactive Learning: Encourage the use of games and puzzles to practice these concepts.
  • Incremental Difficulty: Start with simple problems and gradually increase difficulty as students become more comfortable.
  • Real-world Applications: Show students how these concepts are used in everyday decision-making and scientific research.
Effective mathematics education fosters critical thinking, enabling students to apply mathematical concepts beyond traditional exercises.

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Most popular questions from this chapter

STAYING IN ToucH In a poll conducted in 2007, 2000 adults ages 18 yr and older were asked how frequently they are in touch with their parents by phone. The results of the poll are as follows: $$ \begin{array}{lccccc} \hline \text { Answer } & \text { Monthly } & \text { Weekly } & \text { Daily } & \text { Don't know } & \text { Less } \\ \hline \text { Respondents, \% } & 11 & 47 & 32 & 2 & 8 \\ \hline \end{array} $$ If a person who participated in the poll is selected at random, what is the probability that the person said he or she kept in touch with his or her parents a. Once a week? b. At least once a week?

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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 king of diamonds is drawn.

In a survey conducted in 2007 of 1402 workers 18 yr and older regarding their opinion on retirement benefits, the following data were obtained: 827 said that it was better to have excellent retirement benefits with a lower-than-expected salary, 477 said that it was better to have a higher-than-expected salary with poor retirement benefits, 42 said "neither," and 56 said "not sure." If a worker in the survey is selected at random, what is the probability that he or she answered that it was better to have a. Excellent retirement benefits with a lower-than-expected salary? b. A higher-than-expected salary with poor retirement benefits?

In a survey conducted in the fall 2006, 800 homeowners were asked about their expectations regarding the value of their home in the next few years; the results of the survey are summarized below: $$ \begin{array}{lc} \hline \text { Expectations } & \text { Homeowners } \\ \hline \text { Decrease } & 48 \\ \hline \text { Stay the same } & 152 \\ \hline \text { Increase less than } 5 \% & 232 \\ \hline \text { Increase 5-10\% } & 240 \\ \hline \text { Increase more than 10\% } & 128 \\ \hline \end{array} $$ If a homeowner in the survey is chosen at random, what is the probability that he or she expected his or her home to a. Stay the same or decrease in value in the next few years? b. Increase \(5 \%\) or more in value in the next few years?
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