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Chapter 5: Problem 34

In how many ways can 15 students be lined up?

Short Answer

Expert verified
1,307,674,368,000

Step by step solution

01

Understand the Problem

Determine how many ways 15 students can be arranged in a line.
02

Identify the Formula

The number of ways to arrange a set of objects in a particular order is given by the factorial of the number of objects, denoted as \( n! \). Here, we need to find the factorial of 15, represented as \( 15! \).
03

Calculate 15!

Calculate \( 15! \) using the formula for factorial: \[ 15! = 15 \times 14 \times 13 \times ... \times 2 \times 1 \]
04

Perform the Multiplication

Perform the multiplication step-by-step: \( 15 \times 14 = 210 \)\( 210 \times 13 = 2730 \)Continuing this process until the final product is obtained.
05

Final Result

After performing the multiplication, the final result for \( 15! \) is found to be 1,307,674,368,000.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics dealing with combinations and permutations of objects. It involves counting, arranging, and analyzing different groupings and orderings of items. The key components of combinatorics are:
  • Combinations: Grouping objects where the order does not matter.
  • Permutations: Arranging objects where the order matters.
  • Factorials: Multiplying a series of descending natural numbers.
Combinatorics is essential in fields requiring extensive counting and arrangement analysis, such as computer science, probability, and statistics.
Arrangements
In mathematics, arrangements are often referred to as permutations. An arrangement focuses on the different ways items can be ordered.
For instance, the problem of arranging 15 students in a line involves finding all possible line-ups.
These arrangements depend on:
  • The number of items being arranged (in this case, students).
  • The specific order of each arrangement.
The formula to determine the number of ways to arrange 'n' objects is the factorial of 'n,' written as \( n! \).
For 15 students, the number of possible arrangements is given by \( 15! \).
Permutations
Permutations are a specific type of arrangement where the order of items matters. In the problem of lining up 15 students, permutations help us determine how many different ways we can order these students.
The general formula for the number of permutations of 'n' items is: \[ n! = n \times (n - 1) \times (n - 2) \times ... \times 2 \times 1 \]
When we apply this to 15 students, we calculate: \[ 15! = 15 \times 14 \times 13 \times ... \times 2 \times 1 \]
This calculation equals 1,307,674,368,000 different ways to arrange 15 students. Permutations are crucial in understanding problems that involve ordering, sequencing, and arranging objects uniquely.

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Most popular questions from this chapter

According to Internal Revenue Service records, \(6.42 \%\) of all household tax returns are audited. According to the Humane Society, \(39 \%\) of all households own a dog. Assuming dog ownership and audits are independent events, what is the probability a randomly selected household is audited and owns a dog?

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In airline applications, failure of a component can result in catastrophe. As a result, many airline components utilize something called triple modular redundancy. This means that a critical component has two backup components that may be utilized should the initial component fail. Suppose a certain critical airline component has a probability of failure of 0.006 and the system that utilizes the component is part of a triple modular redundancy. (a) Assuming each component's failure/success is independent of the others, what is the probability all three components fail, resulting in disaster for the flight? (b) What is the probability at least one of the components does not fail?

The weather forecast says there is a \(10 \%\) chance of rain on Thursday. Jim wakes up on Thursday and sees overcast skies. Since it has rained for the past three days, he believes that the chance of rain is more likely \(60 \%\) or higher. What method of probability assignment did Jim use?

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