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Four members from a 50-person committee are to be selected randomly to serve as chairperson, vice-chairperson, secretary, and treasurer. The first person selected is the chairperson; the second, the vice-chairperson; the third, the secretary; and the fourth, the treasurer. How many different leadership structures are possible?

Short Answer

Expert verified
11,037,600 different leadership structures are possible.

Step by step solution

01

Determine the Total Number of People

There are 50 people in the committee from which to choose.
02

Calculate the Number of Choices for Chairperson

Since any of the 50 members can be chosen as the chairperson, there are 50 possible choices for this first position.
03

Calculate the Number of Choices for Vice-Chairperson

After selecting the chairperson, there are 49 remaining members. So, there are 49 possible choices for the vice-chairperson.
04

Calculate the Number of Choices for Secretary

After selecting the chairperson and vice-chairperson, 48 members remain. Thus, there are 48 choices for the secretary.
05

Calculate the Number of Choices for Treasurer

After selecting the chairperson, vice-chairperson, and secretary, 47 members remain. So, there are 47 possible choices for the treasurer.
06

Multiply the Number of Choices

Multiply the number of choices for each position to find the total number of different leadership structures possible: \[ 50 \times 49 \times 48 \times 47 \]
07

Calculate the Total Number of Leadership Structures

Perform the multiplication: \[ 50 \times 49 \times 48 \times 47 = 11037600 \]

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Key Concepts

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

Permutation
In combinatorics, a permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement.
When we say 'arrangement', we are concerned with the different ways in which we can order these objects.

For example, if we have a group of 50 people and want to select 4 different people to fill the roles of chairperson, vice-chairperson, secretary, and treasurer, the order in which we select them matters.
This is because each person will have a different role depending on when they are chosen. This scenario is a classic example of a permutation, not a combination.

In general, if we have a set of 'n' elements and want to pick 'k' elements to arrange, the number of permutations is represented by the formula:
\[ P(n, k) = \frac{n!}{(n - k)!} \]
Here, '!' denotes a factorial, which will be discussed in the next section.
Factorial
A factorial is a mathematical operation where we multiply a series of descending natural numbers. It is denoted by an exclamation point (!).

For instance, the factorial of 5 (written as 5!) is calculated as:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
This concept plays a crucial role in permutations and combinations.

In the exercise above, we used factorials implicitly to calculate the number of ways to arrange 50 people into 4 positions.
Multiplying 50 \times 49 \times 48 \times 47 can be thought of as a part of the factorial 50!, where we omitted all numbers from 46 down to 1.
Essentially, we performed a division in our formula:
\[ P(50, 4) = \frac{50!}{(50 - 4)!} = \frac{50!}{46!} \]
This canceled out all terms in the factorial below 47, simplifying our work to the product of just the first four terms.
Probability
Probability is the measure of the likelihood that an event will occur.
Presented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

When dealing with events like selecting committee leaders, we're dealing with discrete probability, where the outcomes are finite and countable.
In our exercise, we did not calculate a probability directly, but rather the number of possible outcomes (leadership structures).

To find the actual probability of a specific leadership structure being selected randomly from all possible structures, we would use:
\[ P(\text{{specific outcome}}) = \frac{1}{\text{{total number of outcomes}}} \]
Given our calculation showed there are 11,037,600 possible leadership structures, the probability of any one specific structure being chosen randomly is very small:
\[ P(\text{{specific structure}}) = \frac{1}{11037600} \]
This demonstrates the vast number of possible permutations in our example.

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Most popular questions from this chapter

A National Ambulatory Medical Care Survey administered by the Centers for Disease Control found that the probability a randomly selected patient visited the doctor for a blood pressure check is \(0.593 .\) The probability a randomly selected patient visited the doctor for urinalysis is 0.064. Can we compute the probability of randomly selecting a patient who visited the doctor for a blood pressure check or urinalysis by adding these probabilities? Why or why not?

The following data represent the number of drivers involved in fatal crashes in the United States in 2013 by day of the week and gender. $$ \begin{array}{lrrr} & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Sunday } & 4143 & 2287 & 6430 \\ \hline \text { Monday } & 3178 & 1705 & 4883 \\ \hline \text { Tuesday } & 3280 & 1739 & 5019 \\ \hline \text { Wednesday } & 3197 & 1729 & 4926 \\ \hline \text { Thursday } & 3389 & 1839 & \mathbf{5 2 2 8} \\ \hline \text { Friday } & 3975 & 2179 & \mathbf{6 1 5 4} \\ \hline \text { Saturday } & 4749 & 2511 & \mathbf{7 2 6 0} \\ \hline \text { Total } & \mathbf{2 5 , 9 1 1} & \mathbf{1 3 , 9 8 9} & \mathbf{3 9 , 9 0 0} \\ \hline \end{array} $$ (a) Among Sunday fatal crashes, what is the probability that a randomly selected fatality is female? (b) Among female fatalities, what is the probability that a randomly selected fatality occurs on Sunday? (c) Are there any days in which a fatality is more likely to be male? That is, is \(P(\) male \(\mid\) Sunday \()\) much different from \(P(\) male \(\mid\) Monday \()\) and so on?

Suppose that a shipment of 120 electronic components contains 4 defective components. To determine whether the shipment should be accepted, a quality- control engineer randomly selects 4 of the components and tests them. If 1 or more of the components is defective, the shipment is rejected. What is the probability that the shipment is rejected?

According to the Sefton Council Password Policy (August 2007), the United Kingdom government recommends the use of 鈥淓nviron passwords with the following format: consonant, vowel, consonant, consonant, vowel, consonant, number, number (for example, pinray45).鈥 (a) Assuming passwords are not case sensitive, how many such passwords are possible (assume that there are 5 vowels and 21 consonants)? (b) How many passwords are possible if they are case sensitive?

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?

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