/*! 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 7 The factorial function occurs of... [FREE SOLUTION] | 91Ó°ÊÓ

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The factorial function occurs often in probability and statistics. For a non- negative integer \(n\), the factorial is denoted \(n\) ! (which is read " \(n\) factorial") and is defined as follows: First, 0! is defined to be 1. Next, if \(n\) is 1 or larger, then \(n\) ! means \(n(n-1)(n-2) \cdots 3 \times\) \(2 \times 1\). Thus \(3 !=3 \times 2 \times 1=6\). Consult the Tech nology Guide to see how to enter the factorial operation on the calculator. In some counting situations, order makes a difference. For example, if we arrange people into a line (first to last), then each different ordering is considered a different arrangement. The number of ways in which you can arrange \(n\) individuals in a line is \(n !\). a. In how many ways can you arrange 5 people in a line? b. How many people will result in more than 1000 possible arrangements for a line? c. Suppose you remember that your four-digit bank card PIN number uses \(7,5,3\), and 1 , but you can't remember in which order they come. How many guesses would you need to ensure that you got the right PIN number? d. There are 52 cards in an ordinary deck of playing cards. How many possible shufflings are there of a deck of cards?

Short Answer

Expert verified
a) 120 ways; b) At least 7 people; c) 24 guesses; d) Approximately \(8.0658 \times 10^{67}\) ways.

Step by step solution

01

Calculate 5! (Part a)

For arranging 5 people in a line, we use the factorial function: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 \]. Calculate:\[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]. Thus, there are 120 ways to arrange 5 people in a line.
02

Determine Minimum n for More than 1000 Arrangements (Part b)

We need to find the smallest non-negative integer \(n\) such that \(n! > 1000\).Calculate and compare factorials:- \(4! = 24\)- \(5! = 120\)- \(6! = 720\)- \(7! = 5040\)Here, \(7! = 5040\) which is the first factorial greater than 1000.Therefore, at least 7 people are needed to have more than 1000 arrangements.
03

Arrange 4 Numbers in All Possible Ways (Part c)

The numbers \(7, 5, 3,\) and \(1\) can be arranged in different ways.Calculate \(4!\) for all possible PIN orderings:\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \].Thus, you would need to try 24 different orders to ensure guessing the correct PIN number.
04

Calculate 52! for Card Shuffling (Part d)

The number of ways to shuffle a deck of 52 cards is given by \(52!\).Calculating \(52!\) directly:This is very large; more precisely, \(52!\approx 8.0658 \times 10^{67}\), representing a vast number of possible shuffles.

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Key Concepts

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

Permutation
In mathematics, a permutation refers to the arrangement of all the members of a set into a particular sequence or order. When the order of arrangement is important, permutations are the perfect tool to use. This is different from combinations, where order does not matter.

Permutations are crucial in solving real-world problems where the sequence of events or objects is significant. For example, if you are arranging people in a line or trying to determine all possible sequences of numbers as in a PIN, you are essentially dealing with permutations.

The number of permutations of a set of items is calculated using the factorial function. For a set containing "n" items, the number of permutations is denoted as "n!", which is read as "n factorial." This expression mathematically represents the product of all positive integers up to "n". For instance, for five people, the factorial is calculated as:
  • 5! = 5 × 4 × 3 × 2 × 1 = 120
In this context, there are 120 different ways to arrange five people in a line. Permutations help us explore possibilities, especially when each unique order matters, like in arranging keys or digits.
Combinatorics
Combinatorics is a branch of mathematics that focuses on counting, arranging, and finding patterns within a finite set of elements. It is a fundamental part of discrete mathematics and is often intertwined with probability and statistics to solve problems that involve counting and measuring likelihoods.

In combinatorics, we often differentiate between permutations and combinations. While permutations are concerned with ordered arrangements, combinations refer to selecting items from a set where the order doesn’t matter. However, in many practical applications, such as school timetabling or meal planning, order typically matters, making permutations more applicable.

Let's dive into a real-world application where order is crucial: think about the different potential shufflings of a deck of 52 cards, which is calculated as 52!. This results in a massive number, expressed approximately as:
  • 52! ≈ 8.0658 × 10^{67}
This staggering result illustrates the vast possibilities present in even a seemingly simple collection of items like a deck of cards. The sheer enormity of possible outcomes showcases the power of combinatorial mathematics in understanding and managing possible arrangements. With combinatorics, we gain insights into different ways to structure, organize, and understand sets and configurations.
Probability and Statistics
Probability and statistics are closely related fields that leverage the concepts of chance and data to analyze and interpret real-life situations. Probability provides a mathematical framework for quantifying the likelihood of events, while statistics involves collecting, analyzing, interpreting, and presenting empirical data.

Central to these areas is the concept of the factorial function, which often arises in probability calculations, particularly when dealing with permutations and combinations. For instance, consider a scenario where you're trying to remember the correct sequence of a four-digit PIN.
  • If your known digits are 7, 5, 3, and 1, you'd rely on probability to realize there are 4! possible ways (24 arrangements) to position these numbers.
Knowing this helps in determining the probability of randomly guessing the correct sequence in one try, which is 1 out of the 24 possible sequences.

Statistical methods use such principles to draw inferences about a larger population based on sampled data. From predicting outcomes to estimating probabilities or even determining error margins, combinations and permutations play crucial roles. Understanding these mathematical constructs aids in making informed decisions based on data-driven insights, thus enhancing our comprehension of randomness and order in the world around us.

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Most popular questions from this chapter

The background for this exercise can be found in Exercises 11, 12, 13, and 14 in Section 1.4. A manufacturer of widgets has fixed costs of \(\$ 1200\) per month, and the variable cost is \(\$ 40\) per widget (so it costs \(\$ 40\) to produce 1 widget). Let \(N\) be the number of widgets produced in a month. a. Find a formula for the manufacturer's total cost \(C\) as a function of \(N\). b. The highest price \(p\), in dollars, of a widget at which \(N\) widgets can be sold is given by the formula \(p=53-0.01 N\). Using this, find a formula for the total revenue \(R\) as a function of \(N\). c. Use your answers to parts a and \(b\) to find \(a\) formula for the profit \(P\) of this manufacturer as a function of \(N\). d. Use your formula from part c to determine the two break-even points for this manufacturer. Assume here that the manufacturer produces the widgets in blocks of 50 , so a table setup showing \(N\) in multiples of 50 is appropriate. e. Use your formula from part c to determine the production level at which profit is maximized if the manufacturer can produce at most 1500 widgets in a month. As in part d, assume that the manufacturer produces the widgets in blocks of 50 .

The per capita growth rate \(r\) (on an annual basis) of a population of grazing animals is a function of \(V\), the amount of vegetation available. A positive value of \(r\) means that the population is growing, whereas a negative value of \(r\) means that the population is declining. For the red kangaroo of Australia, the relationship has been given \({ }^{22}\) as $$ r=0.4-2 e^{-0.008 v} $$ Here \(V\) is the vegetation biomass, measured in pounds per acre. a. Draw a graph of \(r\) versus \(V\). Include vegetation biomass levels up to 1000 pounds per acre. b. The population size will be stable if the per capita growth rate is zero. At what vegetation level will the population size be stable?

The length \(L\), in inches, of a certain flatfish is given by the formula $$ L=15-19 \times 0.6^{t} \text {, } $$ and its weight \(W\), in pounds, is given by the formula $$ W=\left(1-1.3 \times 0.6^{t}\right)^{3} $$ Here \(t\) is the age of the fish, in years, and both formulas are valid from the age of 1 year. a. Make a graph of the length of the fish against its age, covering ages 1 to 8 . b. To what limiting length does the fish grow? At what age does it reach \(90 \%\) of this length? c. Make a graph of the weight of the fish against its age, covering ages 1 to 8 . d. To what limiting weight does the fish grow? At what age does it reach \(90 \%\) of this weight? e. One of the graphs you made in parts a and c should have an inflection point, whereas the other is always concave down. Identify which is which, and explain in practical terms what this means. Include in your explanation the approximate location of the inflection point.

Recall that the APR (the annual percentage rate) is the percentage rate on a loan that the Truth in Lending Act requires lending institutions to report on loan agreements. It does not tell directly what the interest rate really is. If you borrow money for 1 year and make no payments, then in order to calculate how much you owe at the end of the year, you must use another interest rate, the EAR (the effective annual rate), which is not normally reported on loan agreements. The calculation is made by adding the interest indicated by the EAR to the amount borrowed. The relationship between the APR and the EAR depends on how often interest is compounded. If you borrow money at an annual percentage rate APR (as a decimal), and if interest is compounded \(n\) times per year, then the effective annual rate EAR (as a decimal) is given by $$ \mathrm{EAR}=\left(1+\frac{\mathrm{APR}}{n}\right)^{n}-1 $$ For the remainder of this problem, we will assume an APR of \(10 \%\). Thus in the formula above, we would use \(0.1\) in place of APR. a. Would you expect a larger or a smaller EAR if interest is compounded more often? Explain your reasoning. b. Make a table that shows how the EAR depends on the number of compounding periods. Use your table to report the EAR if interest is compounded once each year, monthly, and daily. (Note: The formula will give the EAR as a decimal. You should report your answer as a percent with three decimal places.) c. If you borrow \(\$ 5000\) and make no payments for 1 year, how much will you owe at the end of a year if interest is compounded monthly? If interest is compounded daily? d. If interest is compounded as often as possiblethat is, continuously - then the relationship between APR and EAR is given by $$ \mathrm{EAR}=e^{\mathrm{APR}}-1 \text {. } $$ Again using an APR of 10\%, compare the EAR when the interest is compounded monthly with the EAR when the interest is compounded continuously.

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