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Chapter 2: Problem 31

The College of Science Council has one student representative from each of the five science departments (biology, chemistry, statistics, mathematics, physics). In how many ways can a. Both a council president and a vice president be selected? b. A president, a vice president, and a secretary be selected? c. Two members be selected for the Dean's Council?

Short Answer

Expert verified
a. 20, b. 60, c. 10.

Step by step solution

01

Identify Problem Type

This is a permutation and combination problem where we need to calculate in how many different ways students can be selected for various council positions.
02

Understand Given Data

There are 5 students from the departments of Biology, Chemistry, Statistics, Mathematics, and Physics. We need to find different ways to choose positions from these students.
03

Calculate Ways for President and Vice President (Part a)

For selecting a president and a vice president, the order of selection matters. So, we use permutations. We have 5 options for president and, after selecting one, 4 remaining options for vice president. Hence, the number of ways is: \[5 \times 4 = 20\]
04

Calculate Ways for President, Vice President, and Secretary (Part b)

Here, the order also matters (a permutation problem). First, choose a president (5 choices), then a vice president from the remaining 4, and finally a secretary from the remaining 3. So, the calculation is:\[5 \times 4 \times 3 = 60\]
05

Calculate Ways for Dean's Council Members (Part c)

For selecting 2 members where the order doesn't matter, use combinations. We choose 2 out of 5 students, which is calculated as follows: \[\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10\]

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Key Concepts

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

Council President and Vice President Selection
When selecting a council president and vice president, it's all about the order. In this case, order matters because the roles of president and vice president are distinct. This concept falls under a specific type of permutation problem.

Think of permutations whenever the sequence or arrangement is crucial. Here, we have five students, each from a different department: Biology, Chemistry, Statistics, Mathematics, and Physics.

First, you choose the president from these five. When the president is chosen, one out of the five is selected, leaving four students behind. Then, the vice president can only be chosen from these remaining four.

This would lead to calculations like this:
  • 5 options for president
  • 4 remaining options for vice president
Putting it all together, we calculate the total number of combinations by multiplying these choices: \[5 \times 4 = 20\]
This tells us that there are 20 different ways to select both a president and a vice president here.
Permutation Calculation
Permutations involve calculating how elements can be arranged when the order is significant. For example, selecting a president, vice president, and a secretary which means assigning roles distinctly.

In order of selection: first choose the president, giving 5 possible decisions.
Once the president is appointed, you then have 4 students left to select from for the vice president role.
Finally, to choose a secretary, from the remaining 3 students.

This is again a permutation problem, and the calculation goes as follows:
  • 5 options for President
  • 4 options for Vice President
  • 3 options for Secretary
By multiplying the number of choices at each stage, we determine how many ways the leadership positions can be filled:\[5 \times 4 \times 3 = 60\]
This means there are 60 ways of assigning these three distinct roles, exemplifying the nature of permutations where order matters.
Combination Calculation
In scenarios where the order doesn't matter鈥攍ike selecting members for a committee or council鈥攜ou use combinations. This is the case when choosing two members to join the Dean's Council.

The task here is to pick any two members out of five without worrying about who's picked first. We use the combination formula for these calculations:
  • "n" is the total number of items to choose from (5 students here)
  • "r" is the number of items to choose (in this case, 2 members)
With these, the formula for a combination becomes: \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]

Plugging in the numbers:\[\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10\]
This indicates that there are 10 ways to choose two students from five, highlighting the essence of combinations where order is irrelevant.

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Most popular questions from this chapter

Four graduating seniors, \(A, B, C\), and \(D\), have been scheduled for job interviews at 10 a.m. on Friday, January 13, at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms \(1,2,3\), and 4 , respectively. Unaware of this, the manager's secretary assigns them to the four rooms in a completely random fashion (what else!). What is the probability that a. All four end up in the correct rooms? b. None of the four ends up in the correct room?

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost . Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Manag., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

A professional organization (for statisticians, of course) sells term life insurance and major medical insurance. Of those who have just life insurance, \(70 \%\) will renew next year, and \(80 \%\) of those with only a major medical policy will renew next year. However, \(90 \%\) of policyholders who have both types of policy will renew at least one of them next year. Of the policy holders \(75 \%\) have term life insurance, \(45 \%\) have major medical, and \(20 \%\) have both. a. Calculate the percentage of policyholders that will renew at least one policy next year. b. If a randomly selected policy holder does in fact renew next year, what is the probability that he or she has both life and major medical insurance?

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent. [Hint: First establish a relationship among \(P\left(A^{\prime} \cap B\right), P(B)\), and \(P(A \cap B)\).]

Consider four independent events \(A_{1}, A_{2}, A_{3}\), and \(A_{4}\) and let \(p_{i}=P\left(A_{i}\right)\) for \(i=1,2,3,4\). Express the probability that at least one of these four events occurs in terms of the \(p_{i}\) 's, and do the same for the probability that at least two of the events occur.
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