Section 5.3: Permutations

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Learning Objectives

Calculate the value of factorial expressions
Find the number of permutations of n objects
Find the number of permutations of n objects taken r at a time
Find the number of permutations when some objects are alike

Previously, we studied the Fundamental Counting Principle, which allows us to find the total number of outcomes by multiplying the choices at each step. For example, 3 shirt choices and 2 pants choices give us 3 $\cdot$ 2 = 6 possible outfits.

Now we explore situations where order matters, like arranging students in a line or letters in a word. This leads us to permutations, which count arrangements where the sequence is important. A permutation is an arrangement of objects where changing the order creates a different result. There are many different types of permutations that exist from basic to advanced mathematical models that need to be used for computation. In this section, we will study three types of introductory permutations, each with a distinct formula or method. But, prior to that we need to understand what is called factorial notation.

Introduction to Factorial Notation

When solving permutation problems, you'll notice a specific multiplication pattern that appears repeatedly. Let's examine this pattern by looking in detail at our previous example of the Fundamental Counting Principle.

Example: How many ways can you arrange the letters A, B, C, D?

Solution:

1st position: We can choose any of the 4 letters (A, B, C, or D) → 4 choices
2nd position: After placing one letter, we have 3 remaining letters to choose from → 3 choices
3rd position: After placing two letters, we have 2 remaining letters to choose from → 2 choices
4th position: After placing three letters, only 1 letter remains → 1 choice

By applying the Fundamental Counting Principle and multiplying the number of available choices at each step, we find that the total arrangements equal 4 $\cdot$ 3 $\cdot$ 2 $\cdot$ 1 = 24 different ways.

You can see that this calculation follows a distinct multiplication pattern where we start with the total number of objects (4) and multiply by each consecutive smaller integer until we reach 1. Because this pattern of multiplying a number by every positive integer smaller than itself down to 1 occurs so frequently in mathematics, it has been given a special name and symbol. This pattern is called "factorial" and is written with an exclamation point (!). Below is the definition of the factorial notation.

Definition: Permutation Type #1 - Factorial Notation

Factorial notation $n!$ represents the product of all positive integers from $n$ down to $1$, expressed by the formula:

\[n!=n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot\cdots\cdot3\cdot2\cdot1 \nonumber \]

Read aloud: "$n!$" is pronounced "$n$ factorial."

So, $5!=5\cdot4\cdot3\cdot2\cdot1=120$, and is read: "five factorial"

Also, we define $0!=1$

Example #5.3.1 🤔

How many ways can 8 different books be arranged on a shelf?

✅ Solution:

So, ORDER MATTERS here. Also, since, all books are to be arranged, we call this a 'basic' or 'full' permutation. Here we use the formula $n!$ where $n=8$. Thus, there are $8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=\text{40,320}$ different book arrangements.

Example #5.3.2 🤔

You have 10 favorite songs on your Spotify playlist. How many ways can you play all 10 songs in order?

✅ Solution:

So, ORDER MATTERS here. Also, since, all songs are to be arranged, we call this a 'basic' or 'full' permutation. Here we use the formula $n!$ where $n=10$. Thus, there are $10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=\text{3,628,800}$ different song orders.

Example #5.3.3 🤔

How many different ways can 6 different colored pencils be arranged in a pencil case?

✅ Solution:

So, ORDER MATTERS here. Also, since, all pencils are to be arranged, we call this a 'basic' or 'full' permutation. Here we use the formula $n!$ where $n=6$. Thus, there are $6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720$ different colored pencil arrangements.

Example #5.3.4 🤔

How many distinct arrangements are possible using the letters in "FLORIDA"?

✅ Solution:

So, ORDER MATTERS here and notice that all 7 letters are different. Also, since, all letters are to be arranged, we call this a 'basic' or 'full' permutation. Here we use the formula $n!$ where $n=7$. Thus, there are $7!=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=\text{5,040}$ different 7-letter distinct arrangements.

Examples #5.3.1 thru #5.3.4 each represent a 'basic' or 'full' permutation, where all the objects are distinct and each is being arranged in a specific order. Here is a quick summary with the examples and the formula.

Basic or Full Permutations

An ordered arrangement of ALL distinct objects.
Example: How many ways can you arrange 5 books on a shelf?
Example: How many 5-letter words can be made from the letters in the word TEXAS?
Uses the formula: $n!$

Now, that we are familiar with ordered arrangements of all distinct objects, what if we wanted to only arrange a few of the distinct objects, or less than all in general. Let's look at the following example:

Example: How many ways can you choose a president and a vice-president from a group of 4 different people?

Solution: So, in this case, we are not using all four individuals, since we are only choosing two from the group of four. To better understand this further, let’s give the following names for the group: Amy, Bill, Cassidy, and Daphne. So, let’s look at the case if Amy is president, then there are three different choices left for the vice-president. Now, let’s look at the case if Bill is president, then there are three different choices for the vice president. After considering the other two cases where Cassidy & Daphne are both president, there are a total of 12 different rankings that can occur and is shown in the table below with the first name assumed president and the second name presumed vice-president.

Amy & Bill	Bill & Amy	Cassidy & Amy	Daphne & Amy
Amy & Cassidy	Bill & Cassidy	Cassidy & Bill	Daphne & Bill
Amy & Daphne	Bill & Daphne	Cassidy & Daphne	Daphne & Cassidy

for simplicity,

Notice, for example, we have AB and BA. Since the question was asking about a President and Vice President, in which there is a ranking, it is implied that both AB and BA are two completely different choices, even though they are both the same set of people. Thus, we are counting both outcomes as two different choices. That is because the ORDER MATTERS here. So, the solution is that there are 12 different choices.

So, what if the above problem had either more people or more positions or both? This would take a great amount of time and effort to answer that question. Luckily, there is a formula that can be used to directly compute the total number of permutations without actually listing them as illustrated above. This is our second type of permutation called a 'partial' permutation.

Definition: Permutation Type #2 - Partial Permutation

The arrangement of $r$ distinct objects taken from a larger set of $n$ objects, where $r < n$, is denoted by $_nP_r$ and is given by the formula:

\[ _nP_r=\frac{n!}{(n-r)!} \nonumber \]

Example #5.3.5 🤔

Referring to the example above, how many ways can you choose a president and a vice-president from a group of 4 different people?

✅ Solution:

So, ORDER MATTERS here, because the titles president and vice-president are different positions. Also, since, only $2$ of the $4$ people are to be selected, we call this a 'partial' permutation.

Thus, we will use the 'partial' permutation formula

\[_nP_r=\frac{n!}{(n-r)!}, \nonumber\]

where $n = 4$ and $r = 2$.

\[_4P_2=\frac{4!}{(4-2)!}=\frac{4!}{2!}=\frac{24}{2}=12 \nonumber\]

Thus, there are $12$ different ways you can choose a president and a vice-president.

✅ Alternative Solution:

Here are some other ways to apply and calculate using the same formula from above:

\[_4P_2=\frac{4!}{(4-2)!}=\frac{4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{4\cdot3\cdot\color{red}\cancel{\color{black}2}\cdot\color{red}\cancel{\color{black}1}}{\color{red}\cancel{\color{black}2}\cdot\color{red}\cancel{\color{black}1}}=4\cdot3=12 \nonumber\]

\[_4P_2=\frac{4!}{(4-2)!}=\frac{4!}{2!}=\frac{4\cdot3\cdot2!}{2!}=\frac{4\cdot3\cdot\color{red}\cancel{\color{black}2!}}{\color{red}\cancel{\color{black}2!}}=4\cdot3=12 \nonumber\]

or using a calculator function to get directly to the solution

\[ _4P_2=12 \nonumber\]

or lastly, we can use the Fundamental Counting Principle here; we can think of $2$ 'slots' where we can place any of the $4$ people. The first slot would have $4$ choices and the second slot would then only have $3$ choices to choose from. Applying the Fundamental Counting Principle from the previous section, we have

\[4\cdot3=12 \nonumber \]

Using the Fundamental Counting Principle is by far the easiest of any of the above methods, except maybe for the direct result from a calculator function. For the purposes of this section, the rest of the examples below that use the 'partial' permutation, we will always use the 'partial' permutation formula, since it is similar to a future calculation involving where order does not matter and where the Fundamental Counting Principle cannot be applied.

Example #5.3.6 🤔

There are 7 different books on a desk. How many ways can you arrange 3 of those 7 books on a shelf?

✅ Solution:

So, ORDER MATTERS here. Also, since, only $3$ out of the $7$ books are to be arranged, we call this a 'partial' permutation.

Thus, we will use the 'partial' permutation formula

\[_nP_r=\frac{n!}{(n-r)!}, \nonumber\]

where $n = 7$ and $r = 3$,

\[ _7P_3=\frac{7!}{(7-3)!}=\frac{7!}{4!}=\frac{\text{5,040}}{24}=210 \nonumber \]

There are $210$ different ways the books can be arranged.

Example #5.3.7 🤔

A committee of 9 people needs to elect a president, a vice-president, a secretary, and a treasurer. How many ways can this be done?

✅ Solution:

So, ORDER MATTERS here, because the titles president, vice-president, secretary, and treasurer are different positions. Also, since, only $4$ of the $9$ people are to be selected, we call this a 'partial' permutation.

Thus, we will use the 'partial' permutation formula

\[_nP_r=\frac{n!}{(n-r)!}, \nonumber\]

where $n = 9$ and $r = 4$,

\[ _9P_4=\frac{9!}{(9-4)!}=\frac{9!}{5!}=\frac{\text{362,880}}{120}=\text{3,024} \nonumber \]

There are $\text{3,024}$ different ways you can choose a president, vice-president, secretary, and a treasurer.

Example #5.3.8 🤔

A photographer wants to arrange 3 people from a group of 8 for a portrait. How many different arrangements are possible?

✅ Solution:

So, ORDER MATTERS here. Also, since, only $3$ out of the $8$ people are to be arranged, we call this a 'partial' permutation.

Thus, we will use the 'partial' permutation formula

\[_nP_r=\frac{n!}{(n-r)!}, \nonumber\]

where $n = 8$ and $r = 3$,

\[ _8P_3=\frac{8!}{(8-3)!}=\frac{8!}{5!}=\frac{\text{40,320}}{120}=336 \nonumber \]

There are $336$ different ways the group can be arranged for a portrait.

Example #5.3.9 🤔

In a race with 10 runners, how many different ways can 1^st, 2^nd, and 3^rd place be awarded?

✅ Solution:

So, ORDER MATTERS here. Also, since, only $3$ out of the $10$ runners are to be arranged, we call this a 'partial' permutation.

Thus, we will use the 'partial' permutation formula

\[_nP_r=\frac{n!}{(n-r)!}, \nonumber\]

where $n = 10$ and $r = 3$,

\[ _{10}P_3=\frac{10!}{(10-3)!}=\frac{10!}{7!}=\frac{\text{3,628,800}}{\text{5,040}}=720 \nonumber \]

There are $720$ different ways that 1st, 2nd, and 3rd place can be awarded.

Examples #5.3.5 thru #5.3.9 each represent a 'partial' permutation, where all the objects are distinct and some but not all objects are arranged in a specific order. Here is a quick summary with the examples and the formula.

Partial Permutations

An ordered arrangement of SOME BUT NOT ALL distinct objects.*
Example: How many ways can you arrange 2 from the 5 books on a shelf?
Example: Create a 4-digit passcode with no repetition of digits.
Uses the formula: $_nP_r=\frac{n!}{(n-r)!}$

* You could use an ordered arrangement of ALL distinct objects here. Since $n=r$, it would revert back to the Full Permutations formula.

So, for the previous two types of permutations that we have discussed, notice that any type of arrangements that were involved always had distinct objects, items, people, etc. For our last type of permutation, we will be looking at the case where we arrange all objects with some objects that are identical.

Recall Example #5.3.4, how many distinct arrangements are possible using the letters in "FLORIDA"? That was a 'basic' or 'full' permutation and the result was 7! = 5,040. Notice all letters were used and all 7 letters were distinct. So, what if the question read:

Example: How many distinct arrangements are possible using the letters in "OHIO"? Here we are using all 4 letters, however they are not distinct. There are two repeat letters being used. The letter O is in the word twice. The other letters H and I are only used once. So, let's look at all of the permutations if the last letter was a Z (or the word "OHIZ".

OHIZ	OIZH	HIZO	IZOH
ZHIO	ZIOH	HIOZ	IOZH
OHZI	OZHI	HZIO	IZHO
ZHOI	ZOHI	HOIZ	IOHZ
OIHZ	OZIH	HZOI	IHOZ
ZIHO	ZOIH	HOZI	IHZO

Now, let's replace the letter Z back with the letter O.

OHIO	OIOH	HIOO	IOOH
OHIO	OIOH	HIOO	IOOH
OHOI	OOHI	HOIO	IOHO
OHOI	OOHI	HOIO	IOHO
OIHO	OOIH	HOOI	IHOO
OIHO	OOIH	HOOI	IHOO

As you can see in the table above, some words are exactly the same, because any instance where the two letter 'O's are swapped or switched for each other and the rest of the letters remain in the same position, then the result is an identical word. These identical words are not to be counted as part of the permutation. So, writing only the distinct words in the table above, we get

OHIO	OIOH	HIOO	IOOH
OHOI	OOHI	HOIO	IOHO
OIHO	OOIH	HOOI	IHOO

Thus, there are only 12 different 4-letter distinct arrangements.

So, what if the word had more letters and multiple sets of letters were identical, for example the word "MISSISSIPPI"? This would be nearly impossible to apply the same method we used above, which was tedious enough. Luckily, there is a formula that can be used to directly compute the total number of permutations without actually listing them as illustrated above. This is our third and last type of permutation called a 'full' permutation with 'identical objects'.

Definition: Permutation Type #3 - Full Permutations with Identical Objects

The arrangement of $n$ objects in which $n_1$ objects are alike, $n_2$ objects are alike, etc., and the total number of such arrangements is given by the formula:

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!} \nonumber \]

Example #5.3.10 🤔

How many distinct arrangements are possible using the letters in "OHIO"?

✅ Solution:

So, ORDER MATTERS here. Also, since, all letters are to be arranged and we have repeat or identical letters, we call this a 'full' permutation with 'identical objects'. Here we need to do a letter count:

4 letters total: There are 2 O's, 1 H, and 1 I.

Thus, we will use the 'partial' permutation formula

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}, \nonumber \]

where $n=4, n_1=2, n_2=1, n_3=1$

\[ \frac{4!}{2!\cdot1!\cdot1!}=\frac{24}{2\cdot1\cdot1}=\frac{24}{2}=12 \nonumber \]

There are 12 different 4-letter distinct arrangements using all the letters in the word "OHIO".

Example #5.3.11 🤔

How many distinct arrangements are possible using the letters in "MISSISSIPPI"?

✅ Solution:

11 letters total: There is 1 M, 4 I's, 4 S's, and 2 P's.

Thus, we will use the 'partial' permutation formula

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}, \nonumber \]

where $n=11, n_1=1, n_2=4, n_3=4, n_4=2$

\[ \frac{11!}{1!\cdot4!\cdot4!\cdot2!}=\frac{\text{39,916,800}}{1\cdot24\cdot24\cdot2}=\frac{\text{39,916,800}}{\text{1,152}}=\text{34,650} \nonumber \]

There are 34,650 different 11-letter distinct arrangements using all the letters in the word "MISSISSIPPI".

Example #5.3.12 🤔

How many distinct arrangements are possible using the letters in "CALIFORNIA"?

✅ Solution:

10 letters total: There is 1 C, 2 A's, 1 L, 2 I's, 1 F, 1 O, 1 R, and 1 N's. (To make our calculations easier, we are going to only focus on the duplicate letters, since the single letters will yield a $1!$ in which will not matter in the calculations).

Thus, we will use the 'partial' permutation formula

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}, \nonumber \]

where $n=10, n_1=2, n_2=2$

\[ \frac{10!}{2!\cdot2!}=\frac{\text{3,628,800}}{2\cdot2}=\frac{\text{3,628,800}}{4}=\text{907,200} \nonumber \]

There are 907,200 different 10-letter distinct arrangements using all the letters in the word "CALIFORNIA".

Example #5.3.13 🤔

How many distinct arrangements are possible using the letters in "ABIBLIOPHOBIA"?

✅ Solution:

13 letters total: There is 2 A's, 3 B's, 3 I's, 1 L, 2 O's, 1 P, and 1 H. (To make our calculations easier, we are going to only focus on the duplicate letters, since the single letters will yield a $1!$ in which will not matter in the calculations).

Thus, we will use the 'partial' permutation formula

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}, \nonumber \]

where $n=13, n_1=2, n_2=3, n_3=3, n_4=2$

\[ \frac{13!}{2!\cdot3!\cdot3!\cdot2!}=\frac{\text{6,227,020,800}}{2\cdot6\cdot6\cdot2}=\frac{\text{6,227,020,800}}{144}=\text{43,243,200} \nonumber \]

There are 43,243,200 different 13-letter distinct arrangements using all the letters in the word "ABIBLIOPHOBIA".

Example #5.3.14 🤔

How many ways can you arrange 8 marbles in a row if 3 are red, 3 are blue, and 2 are green?

✅ Solution:

Think as the colored marbles as letters; i.e., RRRBBBGG. So, ORDER MATTERS here. Also, since, all marbles are to be arranged and we have repeat or identical marbles, we call this a 'full' permutation with 'identical objects'. Here we need to do a marble count:

8 marbles total: There are 3 reds, 3 blues, and 2 greens.

Thus, we will use the 'partial' permutation formula

\[ \frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}, \nonumber \]

where $n=8, n_1=3, n_2=3, n_3=2$

\[ \frac{8!}{3!\cdot3!\cdot2!}=\frac{\text{40,320}}{6\cdot6\cdot2}=\frac{\text{40,320}}{72}=560 \nonumber \]

There are 560 different marble arrangements.

Examples #5.2.10 thru #5.2.14 each represent 'full' permutations with 'identical objects'. Here is a quick summary with the examples and the formula.

Full Permutations with Identical Objects

An ordered arrangement of ALL objects with some objects that are identical.
Example: How many 5-letter words can be made from the letters in the word APPLE?
Example: How many 7-letter words can be made from the letters in the word BALLOON?
Example: A parking lot has 7 spaces in a row. If 4 identical red cars and 3 identical blue cars need to be parked, how many different arrangements are possible?
Uses the formula: $\frac{n!}{n_1!\cdot n_2!\cdot n_3!\cdot~\cdots~\cdot n_p!}$

Which Permutation Formula Should You Use? Here is a flow chart to help you distinguish which type of permutation to use.

$Figure 5.3a.png$

Section 5.3: Permutations [In-Class Exercises]

A family of 5 buys Monster Jam tickets. The 5 seats are all in a row. They want to know how many different ways they could sit. How many different seating arrangements are possible?
A student has 8 alphabet magnets, all different letters. How many different 8‑letter arrangements can they make?
A group of 9 parents from a PTA organization needs to elect a President, Vice President, Secretary, and Treasurer. No parent can hold more than one position. How many different ways can these four positions be filled?
An escape room requires players to press three colored buttons in a specific order. The buttons are Red, Blue, Green, Yellow, and Purple. No button can be used twice. How many different 3 button sequences can be made?
Rita has 13 different ornaments and wants to arrange 4 of them on her mantle.
In the 2026 Winter Olympics held in Milano, Italy, there were 12 different men's hockey teams in the tournament. In how many ways can three teams finish with the Gold, Silver, and Bronze medals?
Find the number of distinguishable permutations of the letters in the word FLORIDA.
Find the number of distinguishable permutations of the letters in the word HAWAII.
Find the number of distinguishable permutations of the letters in the word SENSELESSNESSES. (Means several acts, situations, or qualities of being senseless).

Answers: 1) $5!=120$; 2) $\text{8!=40,320}$; 3) $_9P_4=\text{3,024}$; 4) $_5P_3=60$; 5) $_{13}P_4=\text{17,160}$; 6) $_{12}P_3=\text{1,320}$; 7) $7!=\text{40,320}$; 8) $\frac{6!}{2!\cdot2!}=\frac{720}{2\cdot2}=\frac{720}{4}=180$; 9) $\frac{15!}{7!\cdot5!\cdot2!}=\frac{\text{1,307,674,368,000}}{\text{5,040}\cdot120\cdot2}=\frac{\text{1,307,674,368,000}}{\text{1,209,600}}=\text{1,081,080}$.

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Section 5.3: Permutations [In-Class Exercises]