# 5.3. Permutations

# Permutations

In a previous example, 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.

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

Example 5.3.1

*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 · 3 · 2 = 24 different arrangements.

Example 5.3.2

**Solution**

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 · 5 · 4 · 3 · 2 · 1 · 3 = 1440.

Example 5.3.3

*A*,

*B*,

*C*,

*D*,

*E*}. Find the following:

**a.**The number of four-letter word sequences.

**b.**The number of three-letter word sequences.

**c.**The number of two-letter word sequences.

**Solution**

**a.**The number of four-letter word sequences is 5 · 4 · 3 · 2 = 120.

**b.**The number of three-letter word sequences is 5 · 4 · 3 = 60.

**c.**The number of two-letter word sequences is 5 · 4 = 20.

*n*objects and we are selecting

*r*objects to form permutations. We refer to this as

**permutations of**

**n****objects taken**

**, and we write it as**

*r*at a time**.**

*n*Pr**a.**The number of four-letter word sequences is 5

*P*4 = 120.

**b.**The number of three-letter word sequences is 5

*P*3 = 60.

**c.**The number of two-letter word sequences is 5

*P*2 = 20.

Before we give a formula for *n*P*r*, we’d like to introduce a symbol that we will use a great deal in this as well as in the next chapter.

**Factorial**: *n*! = *n*(*n *− 1)(*n *− 2)(*n *− 3)··· 3 · 2 · 1.

Where *n* is a natural number.

0! = 1

Now we define *n*P*r*.

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

*n*P*r* = *n*(*n* − 1)(*n *− 2)(*n* − 3)···(*n* − *r* +1), or

*n*P*r* =

Where *n* and *r* are natural numbers.

The reader should become familiar with both formulas and should feel comfortable in applying either.

Example 5.3.4

**a.**6P3

**b.**7P2

**Solution**

*n*and

*r*in each case and solve using the formulas provided.

*P*3 = 6 · 5 · 4 = 120, alternately 6

*P*3 = = = = 120

*P*2 = 7 · 6 = 42, or 7

*P*2 = = = 42

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

Example 5.3.5

**Solution**

*A*,

*B*,

*C*, and

*D*. Further suppose that

*A*and

*B*want to sit together. For the sake of argument, we tie

*A*and

*B*together and treat them as one person. The four people are

*CD*. Since is treated as one person, we have the following possible arrangements:

*A*and

*B*could also be tied in the order

*BA*. And they are:

Example 5.3.6

**Solution**

4 | 3 | 2 | 5 | 4 |

*P*3, and 5 · 4 is 5

*P*2. So the answer can be written as (4

*P*3) (5

*P*2) = 480.

Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5*P*2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4*P*3) (5*P*2). We summarize.

**Permutations**: A permutation of a set of elements is an ordered arrangement where each element is used once.**Factorial**:*n*! =*n*(*n*− 1)(*n*− 2)(*n*− 3)···3 · 2 · 1. Where*n*is a natural number. 0! = 1**Permutations of**:*n*Objects Taken*r*at a Time*n*P*r*=*n*(*n*− 1)(*n*− 2)(*n*− 3)···(*n*−*r*+ 1), or*n*P*r*= . Where*n*and*r*are natural numbers.

# Ci**rcular Permutations and Permutations with Similar Elements**

In this section we will address the following two problems.

- In how many different ways can five people be seated in a circle?
- In how many different ways can the letters of the word MISSISSIPPI be arranged?

The first problem comes under the category of *Circular Permutations*, and the second under *Permutations with Similar Elements*.

**Circular Permutations **

Suppose we have three people named *A*, *B*, and *C*. We have already determined that they can be seated in a straight line in 3! or 6 ways. Our next problem is to see how many ways these people can be seated in a circle. We draw a diagram:

It happens that there are only two ways we can seat three people in a circle. This kind of permutation is called a circular permutation. In such cases, no matter where the first person sits, the permutation is not affected. Each person can shift as many places as they like, and the permutation will not be changed. Imagine the people on a merry-go-round; the rotation of the permutation does not generate a new permutation. So in circular permutations, the first person is considered a place holder, and where he sits does not matter.

**Circular Permutations**: The number of permutations of

*n*elements in a circle is (

*n*− 1)!

Example 5.3.7

**Solution**

1 | 4 | 3 | 2 | 1 |

Example 5.3.8

**Solution**

We again emphasize that the first person can sit anywhere without affecting the permutation. So there is only one choice for the first spot. Suppose a man sat down first. The chair next to it must belong to a woman, and there are 4 choices. The next chair belongs to a man, so there are three choices and so on. We list the choices below.

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

So the answer is 144.

Now we address the second problem.

**Permutations with Similar Elements**

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all of the letters different by labeling the letters as follows.

E_{1}LE_{2}ME_{3}NT

Since all the letters are now different, there are 7! different permutations.

Let us now look at one such permutation, say:

LE_{1}ME_{2}NE_{3}T

Suppose we form new permutations from this arrangement by only moving the E’s. Clearly, there are 3! or 6 such arrangements. We list them below:

LE_{1}ME_{2}NE_{3}T

LE_{1}ME_{3}NE_{2}T

LE_{2}ME_{1}NE_{3}T

LE_{3}ME_{3}NE_{1}T

LE_{3}ME_{2}NE_{1}T

LE_{3}ME_{1}NE_{2}T

Because the E’s are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.

Let us suppose there are *n* different permutations of the letters ELEMENT. Then there are *n *· 3! permutations of the letters E_{1}LE_{2}ME_{3}NT. But we know there are 7! permutations of the letters E_{1}LE_{2}ME_{3}NT. Therefore: *n *· 3! = 7!

Or *n* = .

This gives us the method we are looking for.

**Permutations with Similar Elements**:

The number of permutations of *n* elements taken *n* at a time, with *r*_{1} elements of one kind, *r*_{2} elements of another kind, and so on, is

Example 5.3.9

**Solution**

Example 5.3.10

**Solution**

Example 5.3.11

**Solution**

Example 5.3.12

**Solution**

We summarize:

**Circular Permutations**: The number of permutations of*n*elements in a circle is (*n*− 1)!**Permutations with Similar Elements**: The number of permutations of*n*elements taken*n*at a time, with*r*_{1}elements of one kind,*r*_{2}elements of another kind, and so on, such that*n*=*r*_{1}+*r*_{2}+···+*r*_{k}is

This is also referred to as **ordered partitions**.

# Practice questions

**1.** A group of 15 people who are members of a volunteer club wish to choose a chair and a secretary. How many different ways can this be done?

**2.** How many permutations of the letters of the word SECURITY end in a consonant?

**3.** In how many different ways can five people be seated in a row if two of them insist on sitting next to each other?

**4.** In how many ways can 3 English, 3 history, and 2 math books be set on a shelf, if the English books are set on the left, history books in the middle, and math books on the right?

**5.** Find the number of different permutations of the letters of the word MASSACHUSETTS.

**6.** If a team plays 10 games, how many different outcomes of 6 wins, and 4 losses are possible?

Answers

**1. ** 210

**2. ** 25,200

**3. ** 48

**4. ** 72

**5. ** 64,864,800

**6.** 210