/*! 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 32 (a) How many distinct permutatio... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) How many distinct permutations can be made from the letters of the word columns? (b) How many of these permutations start with the letter \(m ?\)

Short Answer

Expert verified
a) The total number of distinct permutations for the word 'columns' is 5040. b) The number of these permutations starting with the letter 'm' is 720.

Step by step solution

01

Understanding Permutations

Permutations refer to the arrangement of objects in a specific order. For a word, a permutation would imply an arrangement of its letters. The number of permutations of a word is given by the factorial of the number of letters. The factorial function (denoted by a !) is defined for an integer n, and denoted by n!. It is the product of all positive integers less than or equal to n.
02

Calculate Total Permutations for 'Columns'

The word 'columns' has 7 letters. So, the total number of ways we can arrange these 7 letters is 7! (7 factorial), which is \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \) possibilities.
03

Calculate Permutations starting with 'm'

For the permutations starting with 'm', 'm' occupies the first spot fixed, leaving 6 spots that can be filled by the remaining 6 letters. Therefore, the number of ways we can arrange the remaining 6 letters is 6!, which is \(6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) possibilities.

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.

Factorial
Understanding the factorial concept is essential when dealing with permutations. A **factorial**, represented by an exclamation point (!), is a mathematical function that multiplies a series of descending natural numbers. For any non-negative integer \( n \), \( n! \) equals the product \( n \times (n-1) \times (n-2) \dots \times 2 \times 1 \). This concept simplifies the process of calculating permutations, as it automatically accounts for all possible ways to arrange objects.

For example, if we have a word with 7 unique letters, such as "columns," the number of different permutations is given by \( 7! \), calculated as \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \). This calculation gives all possible arrangements of the letters, taking into account each possible position they can occupy.
Arrangements
Arrangements, or permutations, describe how objects can be ordered or arranged distinctively. This concept is crucial in various contexts, particularly when dealing with a set of items where order significantly matters.

For example, arranging the letters of the word "columns" involves permutations. To compute all possible distinct arrangements of its 7 letters, we use the factorial of 7, which is \( 7! \). This calculation captures each unique sequence that the letters can form.

When focusing on permutations that start with a specific letter, say "m," the problem simplifies slightly. Here, "m" is fixed in the first position, leaving six other letters to be arranged. The number of permutations starting with "m" is determined by arranging these 6 remaining letters, or \( 6! \). Thus, the number of such unique arrangements is \( 720 \).

These concepts help us understand how to tackle problems that involve arranging items in different sequences or orders.
Probability and Statistics
Though permutations primarily deal with arrangements, they connect deeply with probability and statistics. In probability, permutations help calculate the likelihood of different outcomes based on arrangements. Knowing how to calculate permutations becomes a foundational skill in understanding more complex probabilistic models.

An exercise like determining how many words can start with 'm' from "columns" incorporates basic probability principles. Such calculations assume each permutation is equally likely. Therefore, if you understand how many distinct permutations exist, it's straightforward to determine the probability of a specific arrangement occurring. If there are 5040 total permutations and 720 start with 'm,' the probability that a randomly chosen permutation starts with 'm' is \( \frac{720}{5040} \), simplifying to \( \frac{1}{7} \).

Introducing these concepts at an early stage can develop intuition for solving similar statistical problems which require an understanding of likelihood and frequency of specific outcomes.

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

In a poker hand consisting of 5 cards, find the probability of holding (a) 3 aces; (b) 4 hearts and 1 club.

A firm is accustomed to training operators who do certain tasks on a production line. Those operators who attend the training course are known to be able to meet their production quotas \(90 \%\) of the time. New operators who do not take the training course only meet their quotas \(65 \%\). of the time. Fifty percent of new operators attend the course. Given that a new operator meets his production quota, what is the probability that he (or she) attended the program?

A contractor wishes to build 9 houses, each different in design. In how many ways can he place these houses on a street if 6 lots are on one side of the street and 3 lots are on the opposite side?

An automobile manufacturer is concerned about a possible recall of its best- selling four-door sedan. If there were a recall, there is 0.25 probability that a defect is in the brake system, 0.18 in the transmission, 0.17 in the fuel system, and 0.40 in some other area. (a) What is the probability that the defect is the brakes or the fueling system if the probability of defects in both systems simultaneously is \(0.15 ?\) (b) What is the probability that there are no defects in either the brakes or the fueling system?

A certain federal agency employs three consulting firms \((\mathrm{A}, B,\) and \(C)\) with probabilities \(0.40,\) \(0.35,\) and \(0.25,\) respectively. From past experience it is known that the probability of cost overruns for the firms are \(0.05,0.03,\) and \(0.15,\) respectively. Suppose a cost overrun is experienced by the agency. (a) What is the probability that the consulting firm involved is company C? (b) What is the probability that it is company A?

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.