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

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Previously, we studied the Fundamental Counting Principle, which allows us to find the total number of outcomes of multiple independent events by multiplying the choices at each step. For example, 3 shirt choices and 2 pants choices give us 3 \(\cdot\) 2 = 6 possible outfits. In this example, the order of the choices is not significant.

Now we explore situations where order matters, like arranging students in a line or letters in a word. A permutation is an arrangement of objects where changing the order creates a different result. In this section, we will study three ways of counting permutations, each with a distinct formula or method.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 number of possible arrangements equals 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:

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

• "\(n!\)" 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?

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✅ Solution:

Note that the order of each arrangement is significant. 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?

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✅ Solution:

Note that the order of each arrangement is significant. 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?

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✅ Solution:

Note that the order of each arrangement is significant. 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 of the letters in the word "FLORIDA" are possible?

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✅ Solution:

Note that the order of each arrangement is significant. 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, what if we wanted to only arrange a few of the distinct objects? 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? Assume that no candidate can fill both offices.

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. Without loss of generality, assume we choose the president first. There are four candidates for president, and three remaining for vice president, for a total of 4 \(\cdot\) 3 = 12 different ways the two offices may be filled.

Distinct Arrangements of Amy, Bill, Cassidy, & Daphne

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,

AB

BA

CA

DA

AC

BC

CB

DB

AD

BD

CD

DC

 

Since, it is implied that both AB and BA are two completely different choices, even though they are both the same set of people, this means that the ORDER MATTERS here. (In the next section, we will focus on situations where order does not matter.)

So, what if the above problem had either more people, more positions, or both? Rather than counting all possible permutations individually, we may instead use the following formula. This formula gives the number of "partial" permutations of r objects chosen from n objects.

Definition: Permutation Type #2 - Partial Permutation

The number of distinct arrangements 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?

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✅ 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.

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✅ Alternative Solution:

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

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

or

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

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, we will use the partial permutation formula wherever applicable, in order to emphasize that we are counting permutations (as opposed to combinations, discussed in Section 5.4.)

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?

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✅ 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 from among its members. How many ways can this be done?

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✅ 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 from left to right are possible?

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✅ 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?

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✅ 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: How many ways can you 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.

Note that, up to now, we have assumed that all the objects we choose from are distinct from each other. Now, let us consider what happens when some of these objects are identical with each other.

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.

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 used twice, while the 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".

Distinct Arrangements of O, H, I, & Z

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.

All Arrangements of O, H, I, & 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 the order in which the two Os occur is not significant. These identical words are not to be counted as distinct permutations. So, writing only the distinct words in the table above, we get 

Distinct Arrangements of O, H, I, & O

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

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

What if the word had more letters and multiple sets of letters were identical, for example the word "MISSISSIPPI"? As before, there is a formula that can be used to directly compute the total number of permutations without actually listing them as illustrated above. This formula gives the number of distinct permutations of n objects among which some are identical with others; i.e., number of "full permutation with identical objects".

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

The number of distinct arrangements of \(n\) objects of which \(k_1\) objects are alike, \(k_2\) objects are alike, etc., is given by the formula:

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!} \nonumber \]

Example #5.3.10 🤔

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

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✅ 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 "full permutation with identical objects" formula

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!}, \nonumber \]

where \(n=4, k_1=2, k_2=1, k_3=1\)

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

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

Example #5.3.11 🤔

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

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✅ 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:

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

Thus, we will use the "full permutation with identical objects" formula

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!}, \nonumber \]

where \(n=11, k_1=1, k_2=4, k_3=4, k_4=2\)

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

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 of the letters in the word "CALIFORNIA" are possible?

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✅ 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:

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 will ignore any letters that occur only once, since this amounts to deleting the extraneous factors of 1! from the expression).

Thus, we will use the "full permutation with identical objects" formula

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!}, \nonumber \]

where \(n=10, k_1=2, k_2=2\)

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

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 of the letters in the word "ABIBLIOPHOBIA" are possible?

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✅ 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:

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 "full permutation with identical objects" formula

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!}, \nonumber \]

where \(n=13, k_1=2, k_2=3, k_3=3, k_4=2\)

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

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?

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✅ 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 "full permutation with identical objects" formula

\[ \frac{n!}{k_1!\cdot k_2!\cdot k_3!\cdot~\cdots~\cdot k_p!}, \nonumber \]

where \(n=8, k_1=3, k_2=3, k_3=2\)

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

There are 560 different marble arrangements.

Examples #5.3.10 thru #5.3.14 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 some of which are identical with each other.
• 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.

Section 5.3: Permutations [In-Class Exercises]

1. 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?
2. A student has 8 alphabet magnets, all different letters. How many different 8‑letter arrangements can they make?
3. 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?
4. 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?
5. Rita has 13 different ornaments and wants to arrange 4 of them on her mantle. How many ways can she do so?
6. 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?
7. Find the number of distinguishable permutations of the letters in the word WYOMING.
8. Find the number of distinguishable permutations of the letters in the word HAWAII.
9. Find the number of distinguishable permutations of the letters in the word SENSELESSNESSES. (Meaning 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. \(7!=\text{5,040}\)
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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