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Chapter 7: Problem 46

A student is given a reading list of ten books from which he must select two for an outside reading requirement. In how many ways can he make his selections?

Short Answer

Expert verified
The student can choose two books from the reading list in \(\boldsymbol{45}\) different ways.

Step by step solution

01

Identify the parameters

In this problem, we have a total of \(n = 10\) books, and we need to choose \(k = 2\) books from the list.
02

Use the combination formula to find the number of possible selections

Now we can plug our parameters in the combination formula: \[C(10, 2) = \frac{10!}{(10 - 2)!2!}\]
03

Factorial computation

Now, we'll compute the factorials for the numbers: \(10!\), \(8!\), and \(2!\) \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800\) \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\) \(2! = 2 \times 1 = 2\)
04

Calculate the total number of combinations

Plugging the factorial values into the combination formula: \(C(10, 2) = \frac{3,628,800}{40,320 \times 2} = \frac{3,628,800}{80,640} = 45\) So, there are 45 different ways the student can choose two books from the reading list.

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Key Concepts

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

Factorials
In mathematics, factorials are a crucial concept often used in calculations involving permutations and combinations. A factorial, denoted by the exclamation mark (!), is the product of all positive integers up to a given number. For instance, the factorial of 5, written as 5!, is calculated as follows:
  • Start with the number 5.
  • Multiply by every whole number below it down to 1.
So, we have: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
Factorials grow very quickly, and even a small increase in number can result in a substantial increase in the factorial value. In our reading list problem, we calculated:
  • \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800\)
  • \(8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\)
  • \(2! = 2 \times 1 = 2\)
These calculations are pivotal in determining combinations and permutations, as they provide the foundation for further mathematical operations.
Combination Formula
The combination formula is a mathematical way to find the number of possible selections of items. It is particularly useful when the order of selection does not matter. The formula is expressed as:\[ C(n, k) = \frac{n!}{(n-k)! \, k!} \]where:
  • \(n\) is the total number of items.
  • \(k\) is the number of items to choose.
In the context of our exercise, the student needed to choose 2 books out of 10 available. Plugging into the formula, we used:\[ C(10, 2) = \frac{10!}{(8! \, 2!)} \]
This formula takes the total permutations represented by \(n!\) and adjusts it by dividing through by \((n-k)!\) to eliminate unwanted permutations. It then divides by \(k!\) to account for repetitions, as the order in which the selections are made does not affect the outcome.
Number of Combinations
Calculating the number of combinations is about finding how many ways there are to select items from a group, where the order of selection does not matter. Using the combination formula, we managed to solve our problem:
  • The student had 10 books total.
  • He needed to choose 2 books.
From the combination formula \( C(n, k) \), we found:\[ C(10, 2) = \frac{3,628,800}{40,320 \times 2} = \frac{3,628,800}{80,640} = 45 \]
Thus, there are 45 different ways for the student to select 2 books. Such calculations are commonly used in various fields such as statistics, computer science, and everyday decision-making scenarios, aiding in determining potential outcomes without considering the order of elements.

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