# 7.3: Permutations

- Page ID
- 147300

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objectives

In this section you will learn to:

- Perform calculations using factorials.
- Count the number of possible permutations (ordered arrangement) of n items taken r at a time.
- Count the number of possible permutations when there are conditions imposed on the arrangements.

Prerequisite Skills

Before you get started, take this prerequisite quiz.

1. How many three-letter word sequences can be formed using the letters { A, B, C } if no letter is to be repeated?

**Click here to check your answer**-
\(6\) sequences can be formed.

If you missed this problem,

__review Section 7.2__. (Note that this will open in a new window.)

2. A California license plate consists of a number from 1 to 5, then three letters followed by any three digits. Repetition is possible. How many such plates are possible?

**Click here to check your answer**-
\(87,880,000\) different plates are possible.

If you missed this problem,

__review Section 7.2__. (Note that this will open in a new window.)

3. How many different 4-letter radio station call letters can be made if the first letter must be K or W and no letters can be repeated?

**Click here to check your answer**-
\(27,600\) station names can be made.

If you missed this problem,

__review Section 7.2__. (Note that this will open in a new window.)

## Factorials

When working with the multiplication axiom, we will often need to multiply sequential, descending numbers as we did in Example 5.2.5. We have a special notation for that calculation, which we will use a great deal in this as well as in the next chapter.

Definition: Factorial

\(n!\) is read as "n factorial."

\[\mathrm{n} !=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots 3 \cdot 2 \cdot 1\]

where \(n\) is a natural number.

\(0! = 1\)

Example \(\PageIndex{1}\)

Calculate 5!

*Solution*

Using the definition of a factorial, \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\). Multiplying these numbers gives a value of \(120\).

## Permutations

In Example 5.2.6, we were asked to find the word sequences formed by using the letters { A, B, C } if no letter is to be repeated. The tree diagram gave us the following six arrangements.

ABC, ACB, BAC, BCA, CAB, and CBA.

Arrangements like these, where order is important and no element is repeated, are called permutations.

Definition: Permutations

A permutation of a set of elements is an ordered arrangement where each element is used once.

Example \(\PageIndex{2}\)

Use the multiplication axiom to determine how many three-letter word sequences can be formed using the letters { A, B, C, D }.* *

*Solution*

There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.

4 |
3 |
2 |

Applying the multiplication axiom, we get \(4 \cdot 3 \cdot 2 = 24\) different arrangements.

Example \(\PageIndex{3}\)

Use the multiplication axiom to determine how many permutations of the letters of the word ARTICLE have consonants in the first and last positions*. *

*Solution*

In the word ARTICLE, there are 4 consonants.

Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:

4 |
3 |

Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.

So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get

4 |
5 |
4 |
3 |
2 |
1 |
3 |

So the total permutations are \(4 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 = 1440\).

Example \(\PageIndex{4a}\)

Given five letters { A, B, C, D, E }. Find the following:

- The number of four-letter word sequences.
- The number of three-letter word sequences.
- The number of two-letter word sequences.

*Solution*

The problem is easily solved by the multiplication axiom, and answers are as follows:

- The number of four-letter word sequences is \(5 \cdot 4 \cdot 3 \cdot 2 = 120\).
- The number of three-letter word sequences is \(5 \cdot 4 \cdot 3 = 60\).
- The number of two-letter word sequences is \(5 \cdot 4 = 20\).

We often encounter situations where we have a set of n objects and we are selecting r objects to form permutations. We refer to this as **permutations**** of n objects taken r at a time**, and we usually write it as **nPr**.

Definition: nPr

**The Number of Permutations of n Objects Taken r at a Time**

\[\mathrm{nPr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !} \]

or

\[\mathrm{nPr}=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots(\mathrm{n}-\mathrm{r}+1)\]

where \(n\) and \(r\) are natural numbers.

Note that different texts may use different notation for permutations. \(nPr, P(n,r)\), or \(P^n_r\) all represent permutations.

Example \(\PageIndex{4b}\)

Given five letters { A, B, C, D, E }. Use the permutation formula to find the following:

- The number of four-letter word sequences.
- The number of three-letter word sequences.
- The number of two-letter word sequences.

*Solution*

- The number of four-letter word sequences is 5P4 = \( \dfrac{5!}{(5-4)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{1} = 5 \cdot 4 \cdot 3 \cdot 2 = 120\)
- The number of three-letter word sequences is 5P3 = \( \dfrac{5!}{(5-3)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1} = 5 \cdot 4 \cdot 3 = 60\)
- The number of two-letter word sequences is 5P2 = \( \dfrac{5!}{(5-2)!} = \dfrac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 5 \cdot 4 = 20\)

Example \(\PageIndex{5}\)

An auto service station has 6 employees and 3 bays. How many different ways can the employees be placed at the three bays if each bay only gets one employee?

**Solution**

Since we have 6 employees to select from, \(n=6\). Since we are placing only 3 of them at a time, \(r=3\). Therefore we are trying to calculate 6P3.

\( 6 \mathrm{P} 3=\frac{6 !}{(6-3) !}=\frac{6 !}{3 !}=\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 6 \cdot 5 \cdot 4 = 120 \)

Next we consider some more permutation problems to get further insight into these concepts.

Example \(\PageIndex{6}\)

In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other?* *

**Solution**

The four people are \(\boxed{AB}\) CD. Since \(\boxed{AB}\) is treated as one person, we have the following possible arrangements.

\[ \boxed{AB} CD, \boxed{AB} DC, C \boxed{AB}D, D\boxed{AB}C, CD \boxed{AB}, DC\boxed{AB} \nonumber\]

Note that there are six more such permutations because A and B could also be tied in the order BA. And they are

\[ \boxed{BA}CD, \boxed{BA} DC, C\boxed{BA}D, D\boxed{BA}C, CD\boxed{BA}, DC \boxed{BA} \nonumber\]

So altogether there are 12 different permutations.

Let us now do the problem using the multiplication axiom.

After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3! ways. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements

Example \(\PageIndex{7}\)

You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?* *

*Solution*

We first do the problem using the multiplication axiom.

Since the math books go in the first three slots, there are 4 choices for the first slot,

3 choices for the second and 2 choices for the third.

The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below.

4 |
3 |
2 |
5 |
4 |

Therefore, the number of permutations are \(4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480\).

Alternately, we can see that \(4 \cdot 3 \cdot 2\) is really same as 4P3, and \(5 \cdot 4\) is 5P2.

So the answer can be written as (4P3) (5P2) = 480.

Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4P3) (5P2).

We summarize the concepts of this section:

Note

**1. Factorial**

\[\mathrm{n} !=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots 3 \cdot 2 \cdot 1 \nonumber\]

Where \(n\) is a natural number.

\[0! = 1 \nonumber\]

**2. Permutations**

A permutation of a set of elements is an ordered arrangement where each element is used once.

**3. Permutations of n Objects Taken r at a Time**

\[\mathrm{nPr}=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots(\mathrm{n}-\mathrm{r}+1) \nonumber\]

or

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

where \(n\) and \(r\) are natural numbers.