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Chapter 4: Problem 23

The University of Montana ski team has five entrants in a men's downhill ski event. The coach would like the first, second, and third places to go to the team members. In how many ways can the five team entrants achieve first, second, and third places?

Short Answer

Expert verified
There are 60 different ways for the ski team members to achieve first, second, and third places.

Step by step solution

01

Understanding Permutations

When we want to arrange a subset of items from a larger set, such as determining the order of the first, second, and third places from the ski team entrants, we use permutations. Permutations allow us to calculate the different possible orderings.
02

Identifying the Total Number of Entrants and Positions

We have a total of 5 entrants, and we need to determine the number of ways to arrange these entrants in 3 positions: first, second, and third place.
03

Applying the Permutation Formula

The formula for permutations of choosing r elements from a set of n elements is given by: \[ P(n, r) = \frac{n!}{(n-r)!} \] where \( n = 5 \) and \( r = 3 \), so substituting in, we get: \[ P(5, 3) = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} \]
04

Performing the Calculation

Calculate the factorial expression: \[ \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60 \]Thus, there are 60 different ways for the team members to achieve first, second, and third places.

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Key Concepts

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

Factorial
Factorials are an essential concept in mathematics, especially in permutations and combinations. A factorial is a marked exclamation point
  • It is represented as an exclamation mark (!).
  • It implies multiplying a series of descending natural numbers.
  • For example, the factorial of 5 (written 5!) is calculated by multiplying all positive integers up to 5.
Hence, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
If you calculate the factorial of any number, say \(n\), the calculation is made by a chain of multiplications including all integers from 1 to \(n\).It's a fast-growing function. Even a small increase in \(n\) causes a large jump in the value of the factorial. Factorial calculations often appear in problems involving permutations.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination. It encompasses a wide range of problems and techniques that help in determining how many ways or methods there are to arrange or combine different items. In the field of combinatorics, two main types of problems are particularly emphasized:
  • Permutations: These arrangements are concerned with the order of elements being important. In permutations, the sequence of arranging matters.
  • Combinations: This involves choosing items from a set where order does not matter. In combinations, only the selection of items is considered, not the sequence.
Combinatorics provides a foundational understanding needed for more advanced topics in mathematics and computer science. It helps address real-world issues such as scheduling, network theory, and even genetics.
Permutation Formula
The permutation formula is a key tool in combinatorics for situations where order matters. When determining permutations, you choose \(r\) elements from a set of \(n\) elements, and here's how you calculate it:\[P(n, r) = \frac{n!}{(n-r)!}\]
  • \(n\) represents the total number of items.
  • \(r\) is the number of items you are selecting or arranging.

For the ski team example, with 5 entrants and 3 places:
  • The number of different ways to assign these places is calculated using the formula \(P(5, 3)\).
  • This computes as \(\frac{5!}{(5-3)!}\) or \(\frac{120}{2}\), equating to 60 possible orderings.
The permutation formula allows quick calculation of complex ordering possibilities, making it particularly useful in disciplines that require an understanding of sequences, such as cryptography and algorithm design.

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Most popular questions from this chapter

(a) Draw a tree diagram to display all the possible outcomes that can occur when you flip a coin and then toss a die. (b) How many outcomes contain a head and a number greater than \(4 ?\) (c) Probability Extension Assuming the outcomes displayed in the tree diagram are all equally likely, what is the probability that you will get a head and a number greater than 4 when you flip a coin and toss a die?

You roll two fair dice, a green one and a red one. (a) What is the probability of getting a sum of \(7 ?\) (b) What is the probability of getting a sum of \(11 ?\) (c) What is the probability of getting a sum of 7 or \(11 ?\) Are these outcomes mutually exclusive?

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Survey: Sales Approach In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: $$\begin{array}{llcc} \hline & \text { Sale } & \text { No Sale } & \text { Row Total } \\ \hline \text { Aggressive } & 270 & 310 & 580 \\ \text { Passive } & 416 & 164 & 580 \\ \text { Column Total } & 686 & 474 & 1160 \\ \hline \end{array}$$ Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: \(A=\) aggressive approach, \(P a=\) passive approach, \(S=\) sale, \(N=\) no sale. So, \(P(A)\) is the probability that an aggressive approach was used, and so on. (a) Compute \(P(S), P(S | A),\) and \(P(S | P a).\) (b) Are the events \(S=\) sale and \(P a=\) passive approach independent? Explain. (c) Compute \(P(A \text { and } S \text { ) and } P(P a \text { and } S).\) (d) Compute \(P(N)\) and \(P(N | A).\) (e) Are the events \(N=\) no sale and \(A=\) aggressive approach independent? Explain. (f) Compute \(P(A \text { or } S).\)

Brain Teasers Assume \(A\) and \(B\) are events such that \(0
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