/*! 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 Consider the letters in the word... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the letters in the word LOGARITHM. a. In how many different ways could you select a vowel or a consonant? b. In how many different ways could you select a vowel and then a consonant? c. How many different three-letter "words" (for example, "ORL," "HLG," and "AOI') could you make using each letter no more than once in any one word? (There are three events: "select the first letter," "select the second letter," and "select the third letter." Find the number of ways each event can occur, and then figure out what to do with the three results.)

Short Answer

Expert verified
a. 9 ways, b. 18 ways, c. 504 different three-letter 'words'

Step by step solution

01

Counting vowels and consonants

Determine the number of vowels and the number of consonants in the word LOGARITHM. There are 3 vowels (O, A, I) and 6 consonants (L, G, R, T, H, M).
02

Calculating selection of vowel or consonant

The different ways of selecting a vowel or a consonant is simply the sum of the number of vowels and the number of consonants. This is calculated as 3 (vowels) + 6 (consonants) = 9 ways.
03

Determining vowel-consonant selections

To select a vowel followed by a consonant, multiply the number of ways to select each. For a vowel, it's 3 ways, and for a consonant, it's 6 ways. The total number of ways is 3 (vowels) × 6 (consonants) = 18 ways.
04

Selecting first letter for the three-letter 'word'

The first letter can be any of the 9 letters (3 vowels + 6 consonants), so there are 9 possible choices.
05

Selecting second letter for the three-letter 'word'

After selecting the first letter, there are 8 letters left. So, the second letter can be selected in 8 ways.
06

Selecting the third letter for the three-letter 'word'

Once the first two letters are chosen, there are 7 letters left for the third letter, therefore 7 possible choices.
07

Calculating total combinations for three-letter 'words'

Now, combine the number of ways for each event using the multiplication rule for independent events. The total number of different three-letter 'words' that can be formed is 9 × 8 × 7 = 504 ways.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Permutations and Combinations
Understanding the concepts of permutations and combinations is essential for solving many problems in precalculus, particularly those involving arrangement and selection. The key difference between permutations and combinations lies in the importance of order. A permutation is an arrangement of objects where the order is important, while a combination is a selection of objects where order does not matter.

For example, the word 'LOGARITHM' can generate multiple different three-letter 'words' if we consider the order of letters significant. If the order matters, we're dealing with permutations. In the solution given, selecting a vowel followed by a consonant uses the concept of permutations, meaningful since 'AI' differs from 'IA'.

Permutations of a Single Event

If we have n distinct objects and wish to arrange r of them, the number of permutations is given by the formula: pr = n! / (n - r)! where n! is the factorial of n. For a three-letter 'word' from 'LOGARITHM', selecting the first letter can be done in 9 ways (permutations of a single event). Each choice presents a unique permutation.
Factorial Notation
Factorial notation is a mathematical shorthand used extensively in combinatorics, to indicate the product of an integer and all the non-negative integers below it. Represented by an exclamation mark (!), n!, read as 'n factorial', is calculated as n! = n × (n - 1) × ... × 2 × 1. In cases where n is zero, by definition, 0! = 1.

In the solution example, factorial notation is applied to determine permutations. The factorial simplifies the computation of permutations by providing a quick way to multiply a series of descending natural numbers. For instance, when selecting the third letter in a three-letter 'word', after the first two choices, the 7 remaining letters give 7!, although we don't need the full factorial since we're only choosing the next one in line, which has 7 options.
Multiplication Principle
The multiplication principle, also known as the fundamental counting principle, states that if one event can occur in 'm' ways and a second independent event can occur in 'n' ways, the number of ways both events can occur in sequence is m × n. This principle is foundational in combinatorics and is leveraged to solve complex problems by breaking them down into simpler, sequential events.

In the context of our original problem, the multiplication principle is demonstrated in part c. Once the first letter is selected, the second and third letters are chosen independently of the first. Applying the multiplication principle to the sequential choices, we calculate the total number of different three-letter 'words' as the product of the number of choices for each step: 9 ways for the first letter, 8 for the second, and 7 for the third, resulting in 9 × 8 × 7 = 504 different 'words'. Thus, the multiplication principle takes center stage in solving permutation problems that involve multiple independent selections or arrangements.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Multiple-Choice Test Problem 2: Suppose that you are taking your College Board tests. You answer all the questions you know, and have some time left over. So you decide to guess the answers to the rest of the questions. a. Each question is multiple choice with five choices. If you guess at random, what is the probability of getting an answer right? Of getting an answer wrong? b. When the testing service grades your paper, they give you 1 point if the answer is right and subtract \(\frac{1}{4}\) point if the answer is wrong. What is your mathematically expected score on any question for which you guess at random? c. Suppose that, on one question, you can eliminate one choice you know is wrong, and then randomly guess among the other four. What is your mathematically expected score on this question? Surprisingly low, isn't it? d. Calculate your mathematically expected score on a question for which you can eliminate two of the choices and then for which you can eliminate three of the choices. e. Based on your answers, do you think it is worthwhile guessing answers on a multiplechoice test?

Evaluate the number of combinations or permutations two ways: a. Using factorials, as in the examples of this section b. Directly, using the built-in features of your grapher $$50 P_{20}$$

Multiple-Choice Test Problem 1: A short multiple-choice test has four questions. Each question has five choices, exactly one of which is right. Willie Passitt has not studied for the test, so he guesses answers at random a. What is the probability that his answer on a particular question is right? What is the probability that it is wrong? b. Calculate his probabilities of guessing \(0,1,2,3,\) and 4 answers right. c. Perform a calculation that shows that your answers to part b are reasonable. d. Plot the graph of this probability distribution. e. Willie will pass the test if he gets at least three of the four questions right. What is his probability of passing? f. This binomial probability distribution is an example of a function of \(-?\)

Just before each Supreme Court session, each of the nine justices shakes hands with every other justice. How many handshakes take place? (IMAGE CAN'T COPY)

Perfect Solo Problem: Clara Nett plays a musical solo. She is quite good and guesses that her probability of playing any one note right is \(99 \%\) The solo has 60 notes. (A). Find the probability that i. She plays every note right ii. She makes exactly one mistake iii. She makes exactly two mistakes iv. She makes at least two mistakes v. She makes more than two mistakes (B). What must be Clara's probability of getting any one note right if she wants to have a \(95 \%\) probability of getting all 60 notes right?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.