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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 ways for the team members to achieve first, second, and third places.

Step by step solution

01

Understand the Problem

We need to determine in how many ways the team members can take the first, second, and third places in a specific order at a ski event. This problem involves permutations because the order of placement matters.
02

Choose Members for First, Second, and Third Places

Start by choosing one of the five team members for the first place. There are 5 possible choices for first place since there are 5 members.
03

Determine Possibilities for Second Place

After choosing the first place, only 4 team members are left for the second place. Thus, there are 4 choices for who can be in second place.
04

Calculate Options for Third Place

Once the first and second places are filled, only 3 team members remain eligible for the third place. Therefore, there are 3 possible choices for third place.
05

Calculate Total Permutations

To find the total number of ways to assign the three top places, multiply the number of choices for each position: 5 (first place) \( \times \) 4 (second place) \( \times \) 3 (third place). The formula for permutations when choosing r items from n is \( P(n, r) = n! / (n-r)! \) where \( n = 5 \) and \( r = 3 \).
06

Compute the Result

Compute the multiplication: \[ 5 \times 4 \times 3 = 60 \]So, there are 60 different ways the team members can achieve the top three positions.

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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 focused on counting, arranging, and analyzing configurations of objects. It's a fundamental part of discrete mathematics, providing essential tools for solving counting problems. In the context of permutations, combinatorics helps us determine how many different ways we can order a set of objects. This is important in scenarios where the order of objects is crucial, such as arranging people in a line or assigning prizes.
  • Combinatorics covers a range of topics including combinations and permutations.
  • Combinations focus on selecting objects without concern for order.
  • Permutations, on the other hand, care deeply about the order of arrangement.
The exercise involving team members competing for ski placements is a great example of permutations since the order in which team members finish matters greatly. This highlights the power of combinatorics in streamlining real-world arrangements.
Factorial
Factorial is a key concept in permutations and combinatorics. Represented by the symbol "!", a factorial of a non-negative integer n, denoted as \( n! \), is the product of all positive integers less than or equal to n. Factorials are used to calculate the number of ways to arrange n distinct objects into a sequence.
  • For instance, the factorial of 5, written as \( 5! \), is calculated as \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
  • In permutations, factorial helps in determining how many total arrangements are possible when some positions are filled.
In our skiing example, factorial is applied in figuring out the number of ways to select 3 positions from 5 entrants. The formula \( P(n, r) = n! / (n-r)! \) utilizes factorials to simplify and find the outcome quickly by reducing calculations into predefined expressions.
Order of Arrangement
The order of arrangement is a crucial component when dealing with permutations. It emphasizes that every arrangement is unique if the sequence of arrangement changes, even with the same objects. This makes order critical in determining permutations.
  • Unlike combinations where order is irrelevant, in permutations, changing the position of even one object results in a different permutation.
  • This is why, in our ski event problem, correctly arranging the positions is necessary to identify how many different medal combinations are possible for the team.
Understanding order of arrangement makes it clear why the final result of 60 ways is reached. We meticulously choose one ski member for first place, then another for the subsequent position, ensuring the sequence to reflect the actual finish order. Thus, focusing on the order amplifies the importance of arrangement in solving permutation problems.

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Most popular questions from this chapter

(a) Explain why \(-0.41\) cannot be the probability of some event. (b) Explain why \(1.21\) cannot be the probability of some event. (c) Explain why \(120 \%\) cannot be the probability of some event. (d) Can the number \(0.56\) be the probability of an event? Explain.

There are three nursing positions to be filled at Lilly Hospital. Position 1 is the day nursing supervisor; position 2 is the night nursing supervisor; and position 3 is the nursing coordinator position. There are 15 candidates qualified for all three of the positions. Determine the number of different ways the positions can be filled by these applicants.

The qualified applicant pool for six management trainee positions consists of seven women and five men. (a) How many different groups of applicants can be selected for the positions? (b) How many different groups of trainees would consist entirely of women? (c) Probability extension: If the applicants are equally qualified and the trainee positions are selected by drawing the names at random so that all groups of six are equally likely, what is the probability that the trainee class will consist entirely of women?

During the Computer Daze special promotion, a customer purchasing a computer and printer is given a choice of three free software packages. There are 10 different software packages from which to select. How many different groups of software packages can be selected?

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event \(A\) is the ratio \(\frac{P(A)}{P(n o t A)}=\frac{P(A)}{P\left(A^{c}\right)}\). For instance, if \(P(A)=0.60\), then \(P\left(A^{c}\right)=0.40\) and the odds in favor of \(A\) are \(\frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}\), written as 3 to 2 or \(3: 2 .\) (a) Show that if we are given the odds in favor of event \(A\) as \(n: m\), the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} \cdot\) Hint \(:\) Solve the equation \(\frac{n}{m}=\frac{P(A)}{1-P(A)}\) for \(P(\bar{A})\). (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to \(15 .\) What is the probability of a successful call? (c) A sports announcer says that the odds a basketball player will make a free throw shot are 3 to \(5 .\) What is the probability the player will make the shot?
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