/*! 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 In how many different orders can... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 8

In how many different orders can five runners finish a race if no ties are allowed?

Short Answer

Expert verified
There are 120 different orders in which five runners can finish the race.

Step by step solution

01

Understand the Problem

The problem is asking for the number of different ways five runners can finish a race. Each position must be filled by a different runner, implying no ties are allowed.
02

Use Factorials

To find the number of different orders, use the concept of permutations. The formula for permutations of n distinct objects is given by: \[ P(n) = n! \]Where \( n! \) (n factorial) is the product of all positive integers up to n.
03

Calculate the Factorial

Here, \( n = 5 \). So calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \).This simplifies to: \( 5! = 120 \).

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 91影视!

Key Concepts

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

Understanding Factorials
Factorials are a central concept in permutations and combinatorics. A factorial, represented by an exclamation mark (!), is the product of all positive integers up to a given number. For example, the factorial of 5 (written as 5!) is calculated as follows: 5! = 5 脳 4 脳 3 脳 2 脳 1 = 120.
Factorials grow very quickly. For instance, while 5! equals 120, 6! equals 720. This exponential growth means they are useful for counting large quantities of arrangements or permutations.
Counting the Permutations
Counting permutations involves determining how many ways elements can be arranged in a sequence. The problem in our example is to find how many different orders five runners can finish a race.
Since there are no ties allowed, every possible sequence of the five runners is a permutation.
We use the factorial to calculate this. For 5 runners, the number of permutations is given by 5!, which equals 120. Hence, there are 120 distinct ways the runners can finish the race.
This method can be applied to any number of distinct elements to find the number of possible arrangements.
Introduction to Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arranging, and finding patterns in sets of elements. It encompasses various topics, including permutations, combinations, and the use of factorials.
Permutations, as seen in our example with runners, concern the arrangement of elements where order matters.
Combinatorics also explores combinations, which count the ways to select items from a set where order does not matter. Both concepts are widely used in fields like probability, statistics, and computer science.
By understanding the basics of combinatorics, you can solve a variety of problems involving arrangement and selection.

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

Determine the number of matches played in a single-elimination tournament with n players, where for each game between two players the winner goes on, but the loser is eliminated.

How many terms are there in the expansion of \((x+y+z)^{100} ?\)

In how many different ways can five elements be selected in order from a set with three elements when repetition is allowed?

Give a combinatorial proof that \(\sum_{k=1}^{n} k\left(\begin{array}{c}{n} \\\ {k}\end{array}\right)=n 2^{n-1}\) \([\text {Hint} : \text { Count in two ways the number of ways to select }\) a committee and to then select a leader of the committee. \(]\)

Use the binomial theorem to expand \(\left(3 x^{4}-2 y^{3}\right)^{5}\) into a sum of terms of the form \(c x^{a} y^{b},\) where \(c\) is a real number and \(a\) and \(b\) are nonnegative integers.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
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.