/*! 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 8 A person buying a personal compu... [FREE SOLUTION] | 91Ó°ÊÓ

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A person buying a personal computer system is offered a choice of three models of the basic unit, two models of keyboard, and two models of printer. How many distinct systems can be purchased?

Short Answer

Expert verified
There are 12 distinct systems that can be purchased, as calculated using the counting principle: \( Total\ distinct\ systems = 3 \times 2 \times 2 = 12 \).

Step by step solution

01

Identify the number of options for each component

We are given: - 3 models of the basic unit - 2 models of the keyboard - 2 models of the printer
02

Use the counting principle to find the number of combinations

According to the counting principle, to find the total number of distinct systems that can be purchased, we can multiply the number of options for each component. Total distinct systems = (Number of basic units) × (Number of keyboards) × (Number of printers)
03

Calculate the total number of distinct systems

Now we can plug in the number of options for each component to calculate the total number of distinct systems. Total distinct systems = (3 models of the basic unit) × (2 models of the keyboard) × (2 models of the printer) = 3 × 2 × 2
04

Simplify the expression

To find the final answer, we can simplify the expression: Total distinct systems = 3 × 2 × 2 = 12 Hence, there are 12 distinct systems that can be purchased.

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Key Concepts

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

Counting Principle
Understanding the counting principle is crucial for efficiently solving a wide range of problems in discrete mathematics. This principle, also known as the fundamental counting principle, provides a straightforward method for determining the number of ways multiple independent choices can be made.

When faced with multiple options across different categories, such as choosing a computer system with various models of basic units, keyboards, and printers, you simply multiply the number of choices available in each category. In our example, there are 3 basic unit models, 2 keyboard models, and 2 printer models. Using the counting principle, we calculate the total number of distinct computer system combinations by multiplying these numbers: \( 3 \times 2 \times 2 = 12 \). Hence, there are 12 unique system configurations that a person can purchase.

This principle is foundational in combinatorics, a key area of discrete mathematics, and is often the first step towards understanding more complex concepts like permutations and combinations.
Combinations
Combinations are a fundamental concept in discrete mathematics when it comes to counting and probability. They refer to the different ways of selecting items from a larger set where the order of selection does not matter. For example, if you're picking a team of 3 players from a group of 10, it doesn't matter if you pick player A first or last, the same team is formed.

To calculate combinations, we use the combination formula which is defined for selecting \( k \) items from a set of \( n \) as: \[ C(n, k) = \frac{n!}{k!(n-k)!} \], where \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \). It's important to note that combinations are different from permutations because for permutations, the order of selection is significant.
Permutations
Permutations are another core concept in discrete mathematics that deals with arranging items in a specific order. When the order is important, such as in a race where first, second, and third place are distinct outcomes, permutations are used instead of combinations.

The permutation formula calculates the number of ways \( r \) items can be ordered from a set of \( n \) items and is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \]. For example, if there are 5 books, and we want to know how many ways we can arrange 3 of them on a shelf, we would use permutations because the arrangement (or order) matters.

Understanding when to use permutations versus combinations — order importance being the key difference — is essential when solving many problems in mathematics, computer science, and related fields.

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Most popular questions from this chapter

$$ \text { Expand the expressions in 1-9 using the binomial theorem. } $$ $$ (p-2 q)^{4} $$

Suppose a person offers to play a game with you. In this game, when you draw a card from a standard 52 -card deck, if the card is a face card you win \(\$ 3\), and if the card is anything else you lose \(\$ 1\). If you agree to play the game, what is your expected gain or loss?

Six people attend the theater together and sit in a row with exactly six seats. a. How many ways can they be seated together in the row? b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat? c. Suppose the six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row?

a. If \(p\) is a prime number and \(a\) is a positive integer, how many divisors does \(p^{a}\) have? b. If \(p\) and \(q\) are prime numbers and \(a\) and \(b\) are positive integers, how many possible divisors does \(p^{a} q^{b}\) have? c. If \(p, q\), and \(r\) are prime numbers and \(a, b\), and \(c\) are positive integers, how many possible divisors does \(p^{a} q^{b} r^{c}\) have? d. If \(p_{1}, p_{2}, \ldots, p_{m}\) are prime numbers and \(a_{1}, a_{2}, \ldots, a_{m}\) are positive integers, how many possible divisors does \(p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{m}^{a_{m}}\) have? e. What is the smallest positive integer with exactly 12 divisors?

Use mathematical induction to prove that for all integers \(n \geq 1\), if \(S\) is a set with \(n\) elements, then \(S\) has the same number of subsets with an even number of elements as with an odd number of elements. Use this fact to give a combinatorial argument to justify the identity of exercise 17 .

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