/*! 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 6 There are 8 people in a club. On... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There are 8 people in a club. One person makes up a list of all the possible committees with 2 members. Another pers fer makes up a list tall the possible ble committees with 6 members. True or false: the second list is longer than the first. Explain briefly.

Short Answer

Expert verified
False: Both lists are the same length, each having 28 committees.

Step by step solution

01

Understanding Combinations

To determine how many possible committees can be formed with a given number of people, we use combinations. The formula for combinations is \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n \) represents the total number of people and \( r \) represents the number of people to choose.
02

Calculate the Number of 2-Member Committees

Using the combination formula, find how many committees of 2 members can be formed from 8 people:\[\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28\]There are 28 possible 2-member committees.
03

Calculate the Number of 6-Member Committees

Next, calculate how many 6-member committees can be formed from 8 people:\[\binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8 \times 7}{2 \times 1} = 28\]There are also 28 possible 6-member committees made from the same group of 8 people.
04

Compare the Lengths of the Two Lists

Both lists, the one for 2-member committees and the one for 6-member committees, contain 28 elements. Therefore, neither list is longer than the other.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arranging, and grouping objects. It's particularly useful when you want to determine the number of ways to select or arrange a subset of items from a larger set. In the context of forming committees from a group of people, combinatorics helps us figure out how many different combinations of people can be put together.

To decide how many 2-member or 6-member committees you can make from 8 people, you would use a combinatorial method called combinations. Combinatorics provides tools to systematically and accurately count these possibilities. Understanding combinatorics enables us to tackle problems involving decision-making, organization, and structure.
Binomial Coefficient
The binomial coefficient is represented by the notation \( \binom{n}{r} \), spoken as "n choose r." This is a key concept in combinatorics, used to determine the number of possible combinations of \( r \) items from a set of \( n \) total items.

The binomial coefficient is calculated using the formula:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \),
where \( n! \) is the factorial of \( n \), following the division of factorials logic. This coefficient is crucial for finding how many different groups you can make from a larger group.
  • For example, when choosing any 2 people from a group of 8, the binomial coefficient \( \binom{8}{2} \) calculates to 28, meaning there are 28 possible 2-person groups.
  • Similarly, \( \binom{8}{6} \) will also result in 28, showing that whether you're choosing 2 people or 6 people from a group of 8, the number of combinations remains the same (because choosing 6 is the same as leaving out 2).
Factorial
A factorial, denoted by \( n! \), is a fundamental mathematical operation where you multiply a series of descending natural numbers. When determining combinations, factorials help calculate the number of ways to arrange a subset of objects.

The idea is relatively simple: the factorial of a positive integer \( n \) is the product of all positive integers less than or equal to \( n \). Here's how it works:
  • \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
  • \( 3! = 3 \times 2 \times 1 \)
  • \( 1! = 1 \)
Factorials grow rapidly with larger numbers, which is essential in calculating large-scale combinations or permutations. In the committee example, the factorial formula \( \binom{8}{2} = \frac{8!}{2!(6!)} \) simplifies using these factorial operations, helping to systematically obtain the correct number of combinations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \begin{array}{l}{\text { Of families with } 4 \text { children, what proportion have more girls than boys? You }} \\ {\text { may assume that the sex of a child is determined as if by drawing at random }} \\ {\text { with replacement from }^{2}}\end{array} $$ $$ \lfloor\mathbf{M}][\mathbf{F}] $$ \(\mathbf{M}=\) male, \(\mathrm{F}=\) female

There are 8 people in a club. \(^{3}\) One person makes up a list of all the possible committees with 2 members. Another person makes up a list of all the possible committees with 5 members. True or false: the second list is longer than the first. Explain briefly.

A die will be rolled 10 times. The chance it never lands six can be found by one of the following calculations. Which one, and why? $$ (\mathrm{i})\left(\frac{1}{6}\right)^{10} \quad(\mathrm{ii}) 1-\left(\frac{1}{6}\right)^{10} \quad(\mathrm{iii})\left(\frac{5}{6}\right)^{10} \quad(\mathrm{iv}) 1-\left(\frac{5}{6}\right)^{10} $$

For each question (a-e) below, choose one of the answers (i-viii); explain your choice. Questions A deck of cards is shuffled. What is the chance that- $$\begin{array}{l}{\text { (a) the top card is the king of spades and the bottom card is the queen }} \\ {\text { of spades? }}\end{array}$$ $$\begin{array}{l}{\text { (b) the top card is the king of spades and the bottom card is the king }} \\ {\text { of spades? }}\end{array}$$ $$\begin{array}{l}{\text { (d) the top card is the king of spades or the bottom card is the queen }} \\ {\text { of spades? }}\end{array}$$ $$\begin{array}{l}{\text { (e) of the top and bottom cards, one is the king of spades and the other }} \\ {\text { is the queen of spades? }}\end{array}$$ $$\begin{array}{l}{\text { Answers }} \\ {\text { (i) } 1 / 52 \times 1 / 51} \\\ {\text { (ii) } 1 / 52+1 / 51} \\ {\text { (iii) } 1 / 52 \times 1 / 52} \\\ {\text { (iv) } 1 / 52+1 / 52} \\ {\text { (vi) } 1-(1 / 52 \times 1 / 51)} \\ {\text { (vi) } 1-(1 / 52 \times 1 / 52)} \\ {\text { (vii) } 2 / 52 \times 1 / 51} \\ {\text { (viii) None of the above }}\end{array}$$

It is now generally accepted that cigarette smoking causes heart disease, lung cancer, and many other diseases. However, in the 1950 s, this idea was controversial. There was a strong association between smoking and ill-health, but association is not causation. R. A. Fisher advanced the "constitutional hypothesis:" there is some genetic factor that disposes you both to smoke and to die. To refute Fisher's idea, the epidemiologists used twin studies. They identified sets of smoking-discordant monozygotic twin pairs. "smokingotic" come from one egg and have identical genetic makeup;" smoking-discordant" means that one twin smokes, the other doesn't.) Now there is a race. Which twin dies first, the smoker or the non-smoker? Data from a Finnish twin study are shown at the top of the next page. Data from the Finnish twin study $$ \begin{array}{ccc}{\text { Smokers }} & {\text { Non-smokers }} \\ {\text { All causes }} & {17} & {5} \\ {\text { Coronary heart disease }} & {9} & {0} \\\ {\text { Lung cancer }} & {2} & {0}\end{array} $$ According to the first line of the table, there were 22 smoking-discordant monozygotic twin pairs where at least one twin of the pair died. In 17 cases, the smoker died first; in 5 cases, the non-smoker died first. According to the second line, there were 9 pairs where at least one twin died of coronary heart disease; in all 9 cases, the smoker died first. According to the last line, there were 2 pairs where at least one twin died of lung cancer, and in both pairs the smoker won the race to death. (Lung cancer is a rare disease, even among smokers.) For parts \((a-c)\) , suppose that each twin in the pair is equally likely to die first. so the number of pairs in which the smoker dies first is like the number of heads in coin-tossing. $$\begin{array}{l}{\text { (a) On this basis, what is the chance of having } 17 \text { or more pairs out of } 22} \\ {\text { where the smoker dies first? }}\end{array} $$ $$\begin{array}{l}{\text { (b) Repeat the test in part (a), for the } 9 \text { deaths from coronary heart disease. }} \\ {\text { (c) Repeat the test in part (a), for the } 2 \text { deaths from lung cancer. }} \\ {\text { (d) Can the difference between the death rates for smoking and non- }} \\\ {\text { smoking twins be explained by }}\end{array}$$ $$ \begin{array}{l}{\text { (i) chance? }} \\ {\text { (ii) genetics? }} \\\ {\text { (iii) health effects of smoking? }}\end{array}$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.