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Chapter 1: Problem 16

A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books. How many choices are possible if (a) both books are to be on the same subjcct? (b) the books are to be on different subjects?

Short Answer

Expert verified
(a) There are 42 possible choices if both books are to be on the same subject. (b) There are 94 possible choices if the books are to be on different subjects.

Step by step solution

01

Part (a): Calculating choices for same subject books

Here, we need to find the choices when both books are on the same subject. We will find the number of ways to choose 2 books from each subject and then add them up. 1. Math books: There are a total of 6 math books, so the number of ways to choose 2 math books is: \(C(6, 2) = \frac{6!}{2!(6-2)!} = 15\) 2. Science books: Similarly, there are a total of 7 science books, so the number of ways to choose 2 science books is: \(C(7, 2) = \frac{7!}{2!(7-2)!} = 21\) 3. Economics books: There are a total of 4 economics books, so the number of ways to choose 2 economics books is: \(C(4, 2) = \frac{4!}{2!(4-2)!} = 6\) Now, we will add the choices of math (15), science (21), and economics (6) books: Total choices = 15 + 21 + 6 = 42 Therefore, there are 42 possible choices if both books are to be on the same subject.
02

Part (b): Calculating choices for different subject books

In this part, we need to find the choices when the books are on different subjects. We will find the number of ways to choose 1 book from each pair of subjects, and then add the results. 1. Math and Science: There are a total of 6 math books and 7 science books, so the number of ways to choose 1 math book and 1 science book is: 6 脳 7 = 42 2. Math and Economics: Similarly, there are a total of 6 math books and 4 economics books, so the number of ways to choose 1 math book and 1 economics book is: 6 脳 4 = 24 3. Science and Economics: There are a total of 7 science books and 4 economics books, so the number of ways to choose 1 science book and 1 economics book is: 7 脳 4 = 28 Now, we will add the choices of Math and Science (42), Math and Economics (24), and Science and Economics (28) pairs: Total choices = 42 + 24 + 28 = 94 Therefore, there are 94 possible choices if the books are to be on different subjects.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random events and determining the likelihood of different outcomes. It deals with the calculation of probabilities, events, and outcomes, and is essential in various fields such as finance, insurance, and even everyday decision making.
When we look at the textbook problem about choosing books, probability theory may not be applied directly in the calculation because we are counting the number of ways to choose books rather than calculating the probability of an event. However, combinatorics, which is a related field and important in probability theory, is central to solving this problem. If, for example, we wanted to find the probability of randomly selecting two math books from the collection, we would use probability principles alongside combinatorics to calculate it.
Permutations and Combinations
Permutations and combinations are core concepts in combinatorics, which is a field of mathematics concerning the counting, arrangement, and combination of elements within a set. When we discuss permutations, we're talking about the number of ways to arrange a set of elements where order matters. On the other hand, combinations refer to the number of ways to select elements from a set where order does not matter.

Combination Example in the Textbook Problem

When the student is choosing books from the same subject in the textbook problem, it's a classic example of a combination. Since it doesn't matter in which order the two math books are chosen, we use combinations to find the total number of selections available. This is why we calculate with the formula for combinations, denoted as \(C(n, k)\), which is \(C(n, k) = \frac{n!}{k!(n-k)!}\). The '!' signifies factorial notation, another vital concept in combinatorics.
Factorial Notation
Factorial notation is represented by an exclamation mark (!) and is crucial to understanding combinatorics problems involving permutations and combinations. A factorial represents the product of all positive integers up to a certain number. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
In the context of our textbook problem, factorials enable us to calculate the different combinations of books. For instance, when calculating \(C(6, 2)\), the expression \(6!\) stands for the product of all numbers from 6 down to 1. However, because order doesn't matter when choosing 2 books from 6, we divide by the factorials of 2 (the number of books chosen) and 4 (the difference between the total books and the number chosen), which simplifies our calculations and prevents overcounting.

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Most popular questions from this chapter

Five scparatc awards (best scholarship, best leadership qualities, and so on) are to be presented to selected students from a class of \(30 .\) How many different outcomes are possible if (a) a student can receive any number of awards? (b) each student can receive at most 1 award?

How many outcome sequences are possible when a die is rolled four times, where we say, for instance, that the outcome is 3,4,3,1 if the first roll landed on \(3,\) the sccond on \(4,\) the third on \(3,\) and the fourth on \(1 ?\)

For years, telephone area codes in the United States and Canada consisted of a sequence of three digits. The first digit was an integer between 2 and \(9,\) the second digit was either 0 or \(1,\) and the third digit was any integer from 1 to \(9 .\) How many area codes were possible? How many area codes starting with a 4 were possible?

(a) In how many ways can 3 boys and 3 girls sit in a row? (b) In how many ways can 3 boys and 3 girls sit in a row if the boys and the girls are each to sit together? (c) In how many ways if only the boys must sit together? (d) In how many ways if no two people of the same sex are allowed to sit together?

We have 20 thousand dollars that must be invested among 4 possible opportunities. Each investment must be integral in units of 1 thousand dollars, and there are minimal investments that need to be made if one is to invest in these opportunities. The minimal investments are \(2,2,3,\) and 4 thousand dollars. How many different investment strategies are available if (a) an investment must be made in cach opportunity? (b) investments must be made in at least 3 of the 4 opportunities?
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