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11.2: Permutations

Created by: CK-12

Introduction

The Lock

Telly was sitting in the middle of a whole pile of combination locks when Carey arrived on Tuesday morning.

“What are you doing?” Carey asked sitting next to the pile of locks.

“I am checking to make sure that all of these locks work,” Telly explained. “Here is a pile for you.”

Carey starting working on the locks too. After about five minutes she stopped and looked at Telly.

“You know these shouldn’t even be called combination locks. A lock isn’t a real combination, it’s actually a permutation,” Carey said laughing.

Telly looked at her friend as if she had six heads. She was completely puzzled.

“What do you mean?” Telly asked.

Do you know what Carey means? If you don’t right now, you will by the end of the lesson. This lesson will help you to understand about permutations.

What You Will Learn

By the end of this lesson, you will be able to complete the following skills.

  • Recognize permutations as arrangements in which order is important.
  • Count all permutations of n objects or events.
  • Count all permutations of n objects taken r at a time.
  • Evaluate permutations using permutation notation.

Teaching Time

I. Recognize Permutations as Arrangements in which Order is Important

Order is important in some situations and not important in others. For example, in following a cake recipe, the order in which the events take place is important. You need to crack the eggs before you mix them with the flour. Similarly, you put the icing on the cake only after it has baked.

In buying the ingredients to make a cake, on the other hand, order is not important. Does it matter if you buy the flour before the eggs or the milk before the icing? It doesn’t, so you would say that order is NOT important in buying cake ingredients.

For solving many problems in which order is important, you can use permutations. A permutation is a selection of items in which order is important. To use permutations to solve problems, you need to be able to identify the problems in which order, or the arrangement of items, matters. Let’s look at an example.

Example

Francis Imelda Guzman wants to know how many ways she can arrange her initials, F, I, N. Does order matter for this problem?

Since she wants to arrange each letter in a particular arrangement, the order does matter. This is a permutation not a combination.

Step 1: Write out a single order, or permutation.

FIG

Step 2: Now rearrange those same letters. Did changing the order of the items change the outcome? If so, then order matters.

NGI \ \Longleftarrow different than original

Each arrangement of letters is a different permutation.

II. Count All Permutations of \underline{n} Objects or Events

We can count permutations too. There is a mathematical way to calculate the number of permutations possible given the number of items selected.

To count the number of permutations in a problem you need to look at the problem as a series of choices.

Example

Three taxis – an Acme Cab, a Bluebird Limo, and a Checker – all arrive outside of the BigWig Hotel at exactly the same moment. In how many different ways can the three line up?

One way to look at this problem is as the product of 3 different choices. For choice 1 you can select any of the three cabs, Acme, Bluebird, or Checker.

For choice 2, your options are now limited. You’ve already chosen the first cab, so you now only have 2 cabs to choose from.

Finally, for choice 3 you have only 1 choice left.

You can multiply the three choices together to get the total number of choices, or permutations, as 6.

Here are the 6 different ways the cabs can line up.

&\mathbf{Acme-Bluebird-Checker} && \mathbf{Bluebird-Acme-Checker} && \mathbf{Checker-Acme-Bluebird}\\&\mathbf{Acme-Checker-Bluebird} && \mathbf{Bluebird-Checker-Acme} && \mathbf{Checker-Bluebird-Acme}

Notice that order is important here. Each of the 6 choices, or permutations, is a unique and different arrangement. For example, Acme-Bluebird-Checker is not the same way as Acme-Checker-Bluebird.

What happens when you increase the number of cabs by adding a Decker Taxi? How many different ways are there now?

Starting over, you can now see that there are 4 choices for the first cab, followed by 3 choices, 2 choices, and 1 choice.

The 24 choices are shown below.

& \text{Acme-Bluebird-Checker-Decker} && \text{Bluebird-Acme-Checker-Decker} && \text{Checker-Acme-Bluebird-Decker} && \text{Decker-Acme-Bluebird-Checker}\\& \text{Acme-Bluebird-Decker-Checker} && \text{Bluebird-Acme-Decker-Checker} && \text{Checker-Acme-Decker-Bluebird} && \text{Decker-Acme-Checker-Bluebird}\\& \text{Acme-Checker-Bluebird-Decker} && \text{Bluebird-Checker-Acme-Decker} && \text{Checker-Bluebird-Acme-Decker} && \text{Decker-Bluebird-Acme-Checker}\\& \text{Acme-Checker-Decker-Bluebird} && \text{Bluebird-Checker-Decker-Acme} && \text{Checker-Bluebird-Decker-Acme} && \text{Decker-Bluebird-Checker-Acme}\\& \text{Acme-Decker-Bluebird-Checker} && \text{Bluebird-Decker-Acme-Checker} && \text{Checker-Decker-Acme-Bluebird} && \text{Decker-Checker-Acme-Bluebird}\\& \text{Acme-Decker-Checker-Bluebird} && \text{Bluebird-Decker-Checker-Acme} && \text{Checker-Decker-Bluebird-Acme} && \text{Decker-Checker-Bluebird-Acme}

You can see that when we use mathematics, we can figure out the number of permutations possible.

III. Count Permutations of \underline{n} Objects Taken \underline{r} at a Time

We can find the number of permutations in a group if you include all members of that group. For example, suppose there are 3 cabs in front of a hotel, Acme, Bluebird, and Checker. If all 3 line up to wait for the next customer, the number of different lineups, or permutations, of 3 items taken 3 at a time is:

Again, this is the permutation for three cabs lined up three at a time. We could also say that this is three objects taken three at a time.

What happens when 4 cabs show up at the hotel, but there is only room for 3 cabs to line up? For example, how many different 3-cab lineups would there be if you started with 4 cabs – Acme, Bluebird, Checker, and Decker?

Now, for choice 1 you have four choices instead of 3.

For the choice 2 you have 3 choices instead of 2 and for choice 3 you have 2 choices instead of 1.

The final calculation gives 24 total choices:

This is the answer when we have four options taken three at a time.

Yes. That is true. We do count them down. This is how we can multiply them to find the accurate number of permutations.

IV. Evaluate Permutations Using Permutation Notation

The most efficient way to calculate permutations uses numbers called factorials. Factorials are special numbers that represent the product of a series of descending numbers. The symbol for a factorial is an exclamation sign. Here are some examples.

8! &= 8- \text{factorial} = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\11! &= 11- \text{factorial} = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\4! &= 4- \text{factorial} = 4 \cdot 3 \cdot 2 \cdot 1\\17! &= 17- \text{factorial} = 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1

To compute the values of factorials, simply multiply the series of numbers.

4! &= 4 \cdot 3 \cdot 2 \cdot 1 = 24\\5! &= 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\\8! &= 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 40,320\\11! &= 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 39,916,800

We can use factorials to calculate permutations. Let’s look at an example.

Example

Suppose you have 6 items and you want to know how many arrangements you can make with 4 of the items.

Order matters in this problem, so you need to find the number of permutations there are in 6 items taken 4 at a time.In permutation notation you write the following.

{\color{red}_{6}}P {\color{blue}_{4}} \ \Longleftarrow 6 items taken 4 at a time

In general, permutations are written as:

{\color{red}_{n}}P {\color{blue}_{r}} \ \Longleftarrow {\color{red}n} items taken {\color{blue}r} at a time

To compute {_n}P{_r} you write:

{_{\color{red}n}}P{_{\color{blue}r}}=\frac{{\color{red}n}!}{({\color{red}n}-{\color{blue}r})!}=\frac{\Longleftarrow {\color{red}\text{total items!}}}{\Longleftarrow \ ({\color{red}\text{total items}}-{\color{blue}\text{items taken at a time}})!}

To compute _{6}P _{4} just fill in the numbers:

 {\color{red}_{6}}P {\color{blue}_{4}}&= \frac{{\color{red}6}!}{({\color{red}6}-{\color{blue}4})!}\\{\color{red}_{6}}P {\color{blue}_{2}}&= \frac{{\color{red}6}!}{({\color{red}6}-{\color{blue}2})!}= \frac{6!}{4!}= \frac{6 \times 5 \times \cancel{4 \times 3 \times 2 \times 1}}{\cancel{4 \times 3 \times 2 \times 1}}= \frac{6 \times 5}{1}=30

Notice that it is the product of the values in descending order that tells us how many permutations are possible.

You can use this method to solve any number of permutations.

Example

Find  _{7}P _{3}

Step 1: Understand what  _{7}P _{3} means.

 {\color{red}_{7}}P {\color{blue}_{3}} \ \Longleftarrow 7 items taken 3 at a time

Step 2: Set up the problem.

 {\color{red}_{7}}P {\color{blue}_{3}}= \frac{{\color{red}7}!}{({\color{red}7}-{\color{blue}3})!}= \frac{\Longleftarrow{\color{red} \text{total items!}}}{\Longleftarrow({\color{red} \text{total items}}-{\color{blue} \text{items taken at a time}})!}

Step 3: Fill in the numbers and simplify.

 {\color{red}_{7}}P {\color{blue}_{3}}= \frac{{\color{red}7}!}{({\color{red}7}-{\color{blue}3})!}= \frac{7!}{4!}= \frac{7 \times 6 \times 5 \times \cancel{4 \times 3 \times 2 \times 1}}{\cancel{4 \times 3 \times 2 \times 1}}= \frac{7 \times 6 \times 5}{1}=210

There are 210 possible permutations.

Now let’s go back to the problem from the introduction and apply the information in this lesson.

Real-Life Example Completed

The Lock

Here is the original problem from the introduction. Reread it and then explain what Carey meant when she said the lock shouldn’t be called a combination lock.

Telly was sitting in the middle of a whole pile of combination locks when Carey arrived on Tuesday morning.

“What are you doing?” Carey asked sitting next to the pile of locks.

“I am checking to make sure that all of these locks work,” Telly explained. “Here is a pile for you.”

Carey starting working on the locks too. After about five minutes she stopped and looked at Telly.

“You know these shouldn’t even be called combination locks. A lock isn’t a real combination, it’s actually a permutation,” Carey said laughing.

Telly looked at her friend as if she had six heads. She was completely puzzled.

“What do you mean?” Telly asked.

Remember to explain your thinking.

Solution to Real – Life Example

Carey said the lock should be called a permutation lock because a permutation is a series of numbers in a specific order. If you don’t put the numbers in that order, then the lock will not open. This is not the case with a mathematical combination as you will understand in the next lesson.

Remember, when order is important, it is a permutation.

Vocabulary

Here are the vocabulary words that are found in this lesson.

Permutation
a selection of items in which order is important.
Factorial
a special number that represents the product of a series of descending values.

Time to Practice

Directions: Write whether order is important for each problem. Give an example to explain each answer.

  1. Four different frogs entered the jumping contest – Spots, Dots, Slimey, and Croaky. In how many ways can the 4 finish in first, second, third, and fourth place?
  2. In Scrabble, Denise has the letters A, R, X, O, G, I, and L. How many different 4-letter words can she make?
  3. Six people have signed up to play Scrabble – Tim, Jim, Kim, Pam, Sam, and Cam. Only 4 people can play at one time. How many different 4-player games are possible?
  4. Arnold printed out his 8-page report without putting page numbers on the pages. Now the pages have gotten all mixed up. In how many different ways can Arnold arrange the 8 pages?
  5. The special lunch at Bamboo Restaurant gives you a choice of won-ton or hot-sour soup, a choice of kung pao shrimp or chicken with broccoli, and a choice rice or noodles. How many different special lunches are there?
  6. Rex forgot the password for his ATM bank card. He knows that the password is made of the 4 digits of his birthday, October 24 or 1, 0, 2, 4.? How many different passwords does he need to try to be sure he gets his password?
  7. Javier wrote out wrote out the letters of Jasmine’s name on 7 cupcakes and put them in a box. How many different ways can he take them out of the box one by one?
  8. On the new reality show, Lazybones, the five finalists – Snoozin’ Betty, Lounge Man, The Yawn Meister, Lana Later, and Bob the Procrastinator – compete to see who is the laziest person. On today’s show, the five will be weeded down to 3 super-finalists. How many different ways can the 3 super-finalists be chosen?
  9. Four friends have printed the letters M, E, T, S on the front of their shirts. They’re going to the Mets game and will sit in 4 seats side by side. In how many different ways can the four sit?
  10. How many different 3-digit numbers can you form from the digits 4, 5, and 6?

Directions: Use permutations to solve each problem.

  1. How many 3-digit numbers can Blanche write using the digits 7, 8, and 9 without repeating any of the digits? List the numbers.
  2. How many 4-letter arrangements can Blanche write using the letters A, B, C, D without repeating any of the letters? List the arrangements.
  3. The programmer for Oddball-TV has 5 new half-hour shows she wants to air on Tuesday evenings: Strange Days, Slightly Off, Odd Rod, Bent, and Icky the Great. In how many different orders can she present the shows?
  4. A train has 6 different cars – a passenger car, a baggage car, a mail car, a diner car, a freight car, and a caboose. In how many different orders can the cars be arranged?
  5. The last 5 digits of Beryl’s phone number are 34567. How many numbers have these same 5 digits?
  6. Mia has 7 charms for her charm bracelet – a heart, a moon, a turtle, a cube, a bird, a hoop, and a car. Into how many different orders can she arrange the 7 charms?
  7. One of the charms in Mia’s bracelet in problem 6 above fell off. How many fewer arrangements are there now?
  8. The Soccer Sweepstakes will pay you $700 if you can guess the order of the 6 teams in the Central League. Each guess costs $1. How much would you need to spend to make to make sure you got the correct order?

Directions: Solve each problem.

  1. For her first card trick, Marlo the Magnificent shuffles 4 cards carefully, then has an audience member pull 2 cards out of the 4-card stack, one by one. How many different orders are possible?
  2. For her next trick, Marlo the Magnificent shuffles 5 cards, then has an audience member pull 2 cards out of the 5-card stack, one by one. How many different orders are possible now?
  3. For Marlo’s third trick, she shuffles 5 cards and has an audience member pull 3 cards out one by one. How many different orders are possible now?
  4. Marlo shuffles 6 cards for her final trick and has an audience member pull 4 cards out one by one. How many different orders are possible now?
  5. There are 8 teams in the 8th grade Premier Soccer League. How many different ways can the top 2 teams finish the season?
  6. How many different ways can the top 3 teams in the Premier League finish the season?

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Date Created:

Jan 15, 2013

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Apr 29, 2014
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