/*! 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 3 Seven elementary students are se... [FREE SOLUTION] | 91Ó°ÊÓ

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Seven elementary students are selected to give a 3-minute presentation on what they did during summer vacation. How many different ways can the speakers be arranged?

Short Answer

Expert verified
The students can be arranged in 5040 different ways.

Step by step solution

01

Understand the Problem

The problem asks us to arrange 7 students to give individual presentations. Each arrangement of speakers is considered a different way. This is essentially a permutation problem.
02

Apply Permutation Formula

To find out how many different ways the students can be arranged, we use the permutation formula for arranging 'n' items: \[ n! = n \times (n-1) \times (n-2) \times \, ... \, \times 1 \]For 7 students, it would be 7 factorial (7!), which represents 7 choices for the first position, 6 for the second, 5 for the third, and so on.
03

Calculate 7 Factorial

Calculate 7! step by step:\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \]Start multiplying the numbers:\[ 7 \times 6 = 42 \]\[ 42 \times 5 = 210 \]\[ 210 \times 4 = 840 \]\[ 840 \times 3 = 2520 \]\[ 2520 \times 2 = 5040 \]\[ 5040 \times 1 = 5040 \]
04

Conclusion

The total number of different ways to arrange the 7 students for their presentations is 5040, based on the calculation of 7!.

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Key Concepts

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

Understanding Factorial
The concept of factorial is a cornerstone in permutations and combinations. In mathematics, the factorial of a non-negative integer "n" is denoted by the symbol \( n! \). It represents the product of all positive integers less or equal to "n". For instance, \( 5! \) means \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

This simple concept is very powerful when determining the number of possible arrangements or permutations of a set.
  • Factorials grow rapidly with increasing "n".
  • \( n! \) is crucial in calculating permutations where order matters.
  • The factorial of zero, \( 0! \), is defined as 1 by convention, which often surprises beginners.
Understanding factorials can make solving permutation problems much easier, as it simplifies the process of determining total arrangements.
Arranging Students for Presentations
Arranging students or any group of items is a classic example of applying permutations in real-world scenarios. When a specific order is required, factors such as who goes first or last become critical in determining the number of ways to arrange the group.

In the case of arranging seven students for presentations, we need to think about all possible orders they could present. This context is perfectly suited for using the permutation formula. Each different order of students counts as one unique way to arrange them.

Consider:
  • Each student receives a distinct position in each permutation.
  • Once a student is selected for a position, they don't return to the selection pool for the remaining positions.
  • The factorial function efficiently counts these arrangements.
Mastering the arrangement of students helps students understand broader permutation problems by focusing on individual positioning.
Presentation Order and Permutations
Understanding how to order presentations connects directly to permutation theory. In permutations, the order of items is crucial. Hence, determining the order in which students deliver their presentations is a classic permutation problem.

When we say we are arranging seven students, we mean we are finding the permutations of seven distinct items. Calculating this using factorial notation, like in \( 7! \), gives us the exact number of possible arrangements: 5040 unique orders.

To grasp how permutations impact real-world scenarios like presentations, consider these points:
  • Changing the order of students results in a totally different sequence.
  • Permutations help in task scheduling, organizing events, and even in scenarios requiring fair chances among participants.
  • Grasping this concept sharpens analytical skills necessary for managing and organizing tasks efficiently.
Recognizing the importance of sequence emphasizes how permutations are essential in composing structured, orderly tasks such as presentations.

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Most popular questions from this chapter

What is the range of the values of the probability of an event?

The table below represents the college degrees awarded in a recent academic year by gender. $$ \begin{array}{lccc} & \text { Bachelor's } & \text { Master's } & \text { Doctorate } \\ \hline \text { Men } & 573,079 & 211,381 & 24,341 \\ \text { Women } & 775,424 & 301,264 & 21,683 \end{array} $$ Choose a degree at random. Find the probability that it is a. A bachelor's degree b. A doctorate or a degree awarded to a woman c. A doctorate awarded to a woman d. Not a master's degree

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Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a \(6,\) then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet \(\$ 1\) and won, you could win \(\$ 5 .\) On average, you would win \(\$ 5\) once in 6 rolls and lose \(\$ 1\) on the other 5 rolls - hence the term fair game. In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1 the house might offer 15 to 1 in order to make a profit. Odds can be expressed as a fraction or as a ratio, such as \(\frac{5}{1}, 5: 1,\) or 5 to \(1 .\) Odds are computed in favor of the event or against the event. The formulas for odds are $$ \begin{array}{l} \text { Odds in favor }=\frac{P(E)}{1-P(E)} \\ \text { Odds against }=\frac{P(\bar{E})}{1-P(\bar{E})} \end{array} $$ In the die example, $$ \begin{aligned} &\text { Odds in favor of a } 6=\frac{\frac{1}{6}}{\frac{5}{6}}=\frac{1}{5} \text { or } 1: 5\\\ &\text { Odds against a } 6=\frac{\frac{5}{6}}{\frac{1}{6}}=\frac{5}{1} \text { or } 5: \end{aligned} $$ Find the odds in favor of and against each event. a. Rolling a die and getting a 2 b. Rolling a die and getting an even number c. Drawing a card from a deck and getting a spade d. Drawing a card and getting a red card e. Drawing a card and getting a queen f. Tossing two coins and getting two tails g. Tossing two coins and getting exactly one tail

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