Part 1 Module 5 Factorials, Permutations, and Combinations

Part 1 Module 5Factorials, Permutations, and Combinations

“n factorial”

If n is a positive integer, then n factorial, denoted n!, is the number n multiplied by all the smaller positive integers.

n! = n (n–1) (n–2) … 3 2 1

n! is the number of ways to arrange n objects.

Also,0! = 1

Factorial problems

We want to be able to recognize when a counting problem can be solved directly by using the factorial function, instead of using the Fundamental Counting Principle.

If a counting question asks for the number of ways to arrange (put in order) a collection of n objects, then the number of outcomes is n!

Example: Suppose we have a collection of five books. In how many ways can the five books be arranged in a row?

Answer: The number of ways to arrange five objects in a row is 5! = 120

Example 1

Select the situation where the number of outcomes is 5!

A. We can choose any combination (“all, some, or none”) of pizza toppings from a list of five toppings.

B. We will take a list of five job applicants and rank them from “best” to “worst.”

C. We will form a five-digit number using digits chosen from this set: {2, 4, 5, 7, 9}

D. All of theseE. None of these

Solution 1

A. We can choose any combination (“all, some, or none”) of pizza toppings from a list of five toppings. 25 = 32

B. We will take a list of five job applicants and rank them from “best” to “worst.” 5! = 120

C. We will form a five-digit number using digits chosen from this set: {2, 4, 5, 7, 9} 5 5 5 5 5= 3125

Example 2

The contestants in this year's Ironic Man Octathlon must compete in all of the following events: wolverine shearing; tobacco spitting; manatee racing; squid wrestling; unicycle demolition derby; spaghetti weaving; elephant tipping; cat herding.

The organizers must decide on the order in which the events will be scheduled. How many outcomes are possible?

A. 40,320 B. 256C. 30,240 D. None of these

Solution 2

The contestants in this year's Ironic Man Octathlon must compete in all of the following events: wolverine shearing; tobacco spitting; manatee racing; squid wrestling; unicycle demolition derby; spaghetti weaving; elephant tipping; cat herding.

The organizers must decide on the order in which the events will be scheduled. How many outcomes are possible?

This exercise is asking for the number of ways to arrange eight items.

The number of arrangements or permutations of eight objects is

8! = 40,320

Two special formulas

We are about to encounter two special formulas that can be used to quickly express the number of outcomes for certain kinds of counting problems.

These two formulas are called, respectively, the permutation formula and the combination formula.

The Permutation Formula P(n,r)

Suppose that a counting problem involves selecting and arranging r distinct elements from a set of n elements.

We say that this is a permutation problem, and we say that the number of outcomes is “the number of permutations of n things taken r at a time, abbreviated P(n, r).

€

P(n,r ) =n!

(n − r )!

Example: a Permutation problem

Due to time constraints, the organizers of this year's Ironic Man Octathlon are going scale things back, and produce a pentathlon, instead.

From the original list of eight events (wolverine shearing; tobacco spitting; manatee racing; squid wrestling; unicycle demolition derby; spaghetti weaving; elephant tipping; cat herding), they will select five events and then decide on the order in which the events will be scheduled.

How many outcomes are possible?

Solution to a permutation problem

Due to time constraints, the organizers of this year's Ironic Man Octathlon are going scale things back, and produce a pentathlon, instead. From the original list of eight events (wolverine shearing; tobacco spitting; manatee racing; squid wrestling; unicycle demolition derby; spaghetti weaving; elephant tipping; cat herding), they will select five events and then decide on the order in which the events will be scheduled.


Solution:

Since they are going to select and arrange five distinct elements from a set of eight elements, the number of outcomes is

P(8,5) = 8!/(8 – 5)! = 8!/3! = 40320/6 = 6720 different arrangements.

The Combination Formula C(n,r)

Suppose that a counting problem involves just selecting, but not arranging r distinct elements from a set of n elements.

We say that this is a combination problem, and we say that the number of outcomes is “the number of combinations of n things taken r at a time, abbreviated C(n, r).

€

C(n,r) =n!

(n − r)!r!

“Order doesn’t matter…”

We say that for a combination problem, “order doesn’t matter.”

This means that an outcome is determined entirely by which elements are selected, not by the order in which the elements are selected or listed.

On the other hand, for a permutation problem, order is important.

That means that for a permutation problem, we care not only about which elements are selected, but also about the order in which the elements are selected or arranged.

Example: a combination problem

We have following six coins in a jar: a penny, a nickel, a dime, a quarter, a half-dollar, and a silver dollar.

Two of the coins will be randomly selected and their values will be added.

For example, if we select the dime and the penny, the monetary sum is 11¢. If we select the quarter and the silver dollar, the monetary sum is $1.25.

How many different monetary sums are possible?

Solution: a combination problem

We have following six coins in a jar: a penny, a nickel, a dime, a quarter, a half-dollar, and a silver dollar. Two of the coins will be randomly selected and their values will be added. For example, if we select the dime and the penny, the monetary sum is 11¢. If we select the quarter and the silver dollar, the monetary sum is $1.25. How many different monetary sums are possible?

Solution Note that a monetary sum is determined entirely by which two

coins are selected. The order in which the two coins are selected does not matter.

For example, if we first select the dime and then select the nickel, the result (15¢) is the same as if we first select the nickel and then select the dime.

This is a typical combination problem. The number of outcomes is C(6, 2) = 6!/[(6 – 2)!2!]= 6!/[4!2!] = 720/[24x2] = 720/48 = 15There are 15 different monetary sums.

Example 3

Select the counting problem in which the number of outcomes is P(7, 3).A. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter password with no repeated letters.B. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter sequence which may have repeated characters.C. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter subset.D. All of these.E. None of these.

Solution 3

Select the counting problem in which the number of outcomes is P(7, 3).

We are looking for a problem that involves selecting three distinct elements from a set of seven elements, and arranging the three elements.Consider choice C (which is a wrong answer):C. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter subset.If we are choosing a three-element subset, then we are choosing three distinct elements, but the order in which the three elements are chosen or listed doesn’t matter.For example, this outcome: {a, b, c} is the same as this outcome: {b, c, a}This is combination problem.C(7,3) = 35 different 3-element subsets in a 7-element set.

Solution 3, page 2


We are looking for a problem that involves selecting three distinct elements from a set of seven elements, and arranging the three elements.Consider choice B (which is a wrong answer):B. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter sequence which may have repeated characters.If the password may contain repeated elements, then we aren’t necessarily choosing three distinct elements; we may choose the same element more than once.That means that this isn’t a permutation problem or a combination problem.It is a Fundamental Counting Counting Principle problem:7 7 7= 343 different 3-character passwords.

Solution 3, page 3


We are looking for a problem that involves selecting three distinct elements from a set of seven elements, and arranging the three elements.Consider choice A (which is the correct answer):A. Using letters from the set {a, b, c, d, e, f, g}, form a three-letter password with no repeated letters.If the password may not contain repeated elements, then we are choosing three distinct elements, and since we are arranging them to form a password, order is important.For example, the password abc is different from the password bca.This is a permutation problem where the number of outcomes isP(7,3) = 7!/(7–3)! = 7!/4! = 210

Example 4

Select the counting problem in which the number of outcomes is C(5,3).A. From a group of five waitresses, select three waitresses and tell them that they are fired.B. From a group of five waitresses, select one waitress to bus tables, a second waitress to fold napkins, and a third waitress to polish silverware.C. From a group of five waitresses, assign a waitress to Table #1, a waitress to table #2 and a waitress to table#3; the same waitress may get more than one assignment.D. All of these.E. None of these.

Solution 4

Select the counting problem in which the number of outcomes is C(5,3).We are looking for a problem that involves selecting three distinct people from a group of five people, but all that matters is which three people are chosen; it doesn’t matter who is chosen first or second or third.Consider choice C (which is a wrong answer):C. From a group of five waitresses, assign a waitress to Table #1, a waitress to table #2 and a waitress to table#3; the same waitress may get more than one assignment.In this case, the same waitress may be selected more than once. Since we aren’t necessarily selecting three distinct people, this isn’t a combination problem or a permutation problem. It is a Fundamental Counting Principle problem.We have five options when a choose the waitress for table #1, five options when we choose the waitress for table #2, and five options when we choose the waitress for table #3.5 5 5 =125 different outcomes.

Solution 4, page 2

Select the counting problem in which the number of outcomes is C(5,3).

Consider choice B (which is a wrong answer):B. From a group of five waitresses, select one waitress to bus tables, a second waitress to fold napkins, and a third waitress to polish silverware.In this case, the wording suggests that we are selecting three distinct waitresses, but, since they are being assigned to different tasks, order is important.For example, if Ann buses tables, Beth folds napkins, and Cher polishes silverware, that is different from if Beth folds napkins, Cher buses tables, and Ann polishes silverware. We care not only about which three people are selected, but also about who gets which job.This is a permutation problem:P(5,3) = 5!/(5–3)! = 5!/2! = 60 different outcomes.

Solution 4, page 3

Select the counting problem in which the number of outcomes is C(5,3).

Consider choice A (which is the correct answer):A. From a group of five waitresses, select three waitresses and tell them that they are fired.In this case, we are selecting three distinct people. Furthermore, since the three people are all being treated the same as one another, it doesn’t matter who we pick first or second or third; all that matters is who the three people are.For example, Ann, Beth and Cher get fired, it is the same as if Beth, Cher, and Ann get fired. Order doesn’t matter.This is a combination problem:C(5,3) = 5!/[(5–3)!3!] = 5!/(2!3!) = 10 different outcomes.

Guidelines

When a problem involves selecting individuals from a group:

1. If it is possible to select the same individual more than once, then it isn’t a P(n,r) problem and it isn’t a C(n,r) problem. It is a FCP problem.

Otherwise:2. If the individuals are being treated differently from one other (for instance, if they are being given different jobs, titles or gifts), then order matters and it is a P(n,r) problem (unless one selection involves a special condition, in which case it is FCP).3. If the individuals are being treated the same as one another (for instance, if they are all getting the same job, title or gift), then order doesn’t matter and it is a C(n,r) problem.

Example 5

A group of ten hobbits gets together to compare rings. Prior to the meeting, each hobbit shakes hands with each of the other hobbits. How many handshakes occur?A. 10!B. 100C. 90 D. 45 E. None of these

Solution 5

A group of ten hobbits gets together to compare rings. Prior to the meeting, each hobbit shakes hands with each of the other hobbits. How many handshakes occur?

Suppose you were managing this process. Every time you form a handshake, you do so by selecting two distinct hobbits and telling them to shake hands.However, in forming a handshake, all that matters is which two hobbits are selected; the order in which they are selected or listed doesn’t matter.For example, if Frodo shakes hands with Sam, it is the same as if Sam shakes hands with Frodo.This is a combination problem:C(10,2) = 45 different handshakes.

Example 6

Harpo, Groucho, Chico, Zeppo, Gummo, Karl, and Skid have won three tickets to the opera. They will randomly choose three people from their group to attend the opera. How many outcomes are possible?A. 21B. 35C. 210D. 343E. None of these

Solution 6

Harpo, Groucho, Chico, Zeppo, Gummo, Karl, and Skid have won three tickets to the opera. They will randomly choose three people from their group to attend the opera. How many outcomes are possible?

An outcome is determined by selecting three distinct people from a group of seven people.This means that the number of outcomes is either P(7,3) or C(7,3), depending upon whether or not order matters.Since the three people are all getting the same reward, order doesn’t matter.For example, if Harpo, Groucho and Chico go to the opera, it is the same as if Chico, Groucho, and Harpo go to the opera.All that matters is which three people are selected, not the order in which h they are selected.So, this is a combination problem.C(7,3) = 35 different three-person combinations.

Example 7

Here is a similar problem in which the number of outcomes is P(7,3) instead of of C(7,3).

Harpo, Groucho, Chico, Zeppo, Gummo, Karl, and Skid have won three tickets to the opera. They will randomly choose three people from their group to attend the opera.

One person gets to ride to the opera in a taxi, another gets to ride to the opera in the bed of a turnip truck, and the third person gets to ride in a rickshaw.


Solution 7

One person gets to ride to the opera in a taxi, another gets to ride to the opera in the bed of a turnip truck, and the third person gets to ride in a rickshaw…

Once again an outcome is determined by selecting three distinct people from a group of seven people.This means that the number of outcomes is either P(7,3) or C(7,3), depending upon whether or not order matters.In this case, since the three people are getting different rewards, order is important.For example, if Harpo rides the taxi, Groucho rides in the turnip truck and Chico rides in the rickshaw, it is a different outcome from the case where Chico rides the taxi, Groucho rides the turnip truck, and Harpo rides the rickshaw.So, this is a permutation problem.P(7,3) = 210 different outcomes.

Exercise #8

Select the counting problem where the number of outcomes is C(6,3). Each problem involves making three selections from this group of six people:Homer, Gomer, Homerina, Gomerina, Plato, Mrs. Plato

A. Select a president, vice-president, and treasurer. Nobody may hold more than one office.

B. Select a person to receive a pencil, select a person to receive a calculator, select a person to receive a sheet of scratch paper. It is possible that the same person might be selected to receive more than one of these gifts.

C. Select three people to take the rhinoceros for a walk.

D. Select a president, a vice-president, and a treasurer. Nobody can hold more than one office, but Gomer, who is bad at math, is not eligible to be treasurer.

E. None of these

Solution #8


In a selection process, if it is possible to select the same person or element more than once, then the problem does not involve P(n,r) or C(n,r). Use the Fundamental Counting Principle.

B. Select a person to receive a pencil, select a person to receive a calculator, select a person to receive a sheet of scratch paper. It is possible that the same person might be selected to receive more than one of these gifts.

Since the same person might be selected more than once, this is not a P(n,r) problem or a C(n,r) problem. Use the FCP.Three decisions:Who gets the pencil? 6 optionsWho gets the calculator? 6 optionsWho gets the scratch paper? 6 options6x6x6 = 216 different outcomes.

Solution #8 p. 2


If a counting problem involves selecting a number of distinct people (that is, the same person can’t be selected more than once), then it is either P(n,r) or C(n,r).In this case, if the individuals are being treated the same as one another, then order doesn’t matter and it is a C(n,r) problem.If the individuals are being treated differently from one another (getting different jobs, titles or rewards, for instance), then order is important, and it is a P(n,r) problem.

A. Select a president, vice-president, and treasurer. Nobody may hold more than one office. P(6,3) = 120 different outcomes

C. Select three people to take the rhinoceros for a walk.C(6,3) = 20 different outcomes

Solution #8 p.3

Select the counting problem where the number of outcomes is C(6,3). Each problem involves making three selections from this group of six people:Homer, Gomer, Homerina, Gomerina, Plato, Mrs. PlatoThere is one exception to the previous guideline. If distinct individuals or elements are selected and assigned different tasks et c., but one of the selections involves a special consideration, then P(n,r) will not give the correct answer. Use the FCP, giving first priority to the selection that involves the special consideration.

D. Select a president, a vice-president, and a treasurer. Nobody can hold more than one office, but Gomer, who is bad at math, is not eligible to be treasurer.

Three decisions:Select treasurer - 5 options (treasurer can be anybody but Gomer)Select president - 5 options (anybody but the person already chosen to be treasurer)Select vice-president - 4 options (anybody but the two people already selected to be treasurer and president)5x5x4 = 100 different outcomes.

Exercise #9

There are 8 teams in a recreational beer-pong league. They are going to schedule a round-robin tournament (in which each team will play each of the other teams one time). How many games will be played?

A. 256B. 64C. 56D. 28E. None of these

Solution #9

There are 8 teams in a recreational beer-pong league. They are going to schedule a round-robin tournament (in which each team will play each of the other teams one time). How many games will be played?

Each game is determined by selecting two distinct teams from a list of eight teams. The order in which the two teams are selected or listed doesn’t matter (for instance, if the Bears play the Lions, it is the same as if the Lions play the Bears).This means that the number of outcomes is C(8,2).

C(8,2) = 8!/[(6!)(2!)] =28 The correct choice is C.

Exercise #10

Macbeth is trying to guess the password for Gomerina's email account. He knows that the password consists of 4 letters chosen from this set:{g,o,m,e,r,i,n,a}.How many passwords are possible, if a password does not contain repeated letters and the third letter is a vowel?A. 32B. 256C. 840D. 1344E. None of these

Solution #10

Macbeth is trying to guess the password for Gomerina's email account. He knows that the password consists of 4 letters chosen from this set: {g,o,m,e,r,i,n,a}.How many passwords are possible, if a password does not contain repeated letters and the third letter is a vowel?

In order to guess a password Macbeth has to make four dependent decisions. Since the choice of the third letter affects the other choices, that decision must be made first. There are four options for the third letter, and then there are seven options for the first letter, six options for the second letter, and five options for the last letter.According to the Fundamental Counting Principle, the number of possible outcomes is (4)(7)(6)(5) = 840.

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Part 1 Module 5 Factorials, Permutations, and Combinations