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# 12.6: Permutations

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Order of Entertainment

Greg and Joanna are in charge of creating an order for the Talent Show. There are 6 sixth graders, 10 seventh graders and 6 eighth graders who have entered the contest.

The order that the students perform in makes a difference, and the sixth graders will all perform together. Then the seventh graders will perform and finally the eighth graders will perform.

Greg and Joanna begin with the sixth graders. Since there are six sixth graders who are performing and order does make a difference, how many different arrangements of the order are possible?

“I don’t have a clue how to figure this out,” Joanna says to Greg.

“I think I do, let me get a piece of paper.”

While Greg gets a piece of paper, it is time for you to think about this problem. Solving it has to do with something called a “permutation.” This lesson is all about permutations and how to figure them out.

Pay attention and at the end of the lesson you will know how to help Greg and Joanna figure out the order of sixth graders.

What You Will Learn

By the end of this lesson, you will know how to accomplish the following skills tests:

• Recognize permutations as arrangements in which order is important.
• Count all permutations of n\begin{align*}n\end{align*} objects or events.
• Count permutations of n\begin{align*}n\end{align*} objects taken r\begin{align*}r\end{align*} at a time.
• Evaluate permutations using permutation notation.

Teaching Time

I. Recognize Permutations as Arrangements in which Order is Important

You learned in that last lesson about combinations. You can make all kinds of combinations. Let’s say that you are making a pizza with pepperoni, mushrooms and peppers. It doesn’t matter which order you put the toppings on the pizza. You will still have the same pizza.

Sometimes, order does make a difference. When you arrange different objects or events and order is important, we call each arrangement a permutation. This lesson is all about permutations.

Consider the word CAT. Clearly, order is important when you spell a word. You can write all of the correct letters, but if you don’t put them in the correct order, you don’t spell CAT. For example, here are some orders of C, A, and T that don’t spell CAT;

Incorrect orders:ACT,ATC,CTA,TAC,TCACorrect orders: CAT\begin{align*}\text{Incorrect orders:} \quad \quad ACT, ATC, CTA, TAC, TCA\!\\ \text{Correct orders:} \quad \quad \ CAT\end{align*}

This is an example where a permutation makes a difference. We can also use permutations to solve problems. 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

Tomás wants to know how many 3-digit numbers he can write using the digits 7, 8, and 9 without repeating any of the digits. Does order matter for this problem?

Step 1: Write out a single order.

789

Step 2: Now rearrange the order. Did you change the outcome? If so, then order matters.

798 This is different from the original number

Each arrangement of digits is a different permutation.

Example

The softball coach needs to determine how many different batting lineups she can make out of her first three batters, Able, Baker, and Chan. Does order matter for this problem?

Step 1: Write out a single lineup.

Able, Baker, Chan

Step 2: Now rearrange the lineup. Did you change the outcome? If so, then order matters.

Able, Chan, Baker–this order is different from the original

Each arrangement of batters is a different permutation.

12J. Lesson Exercises

Write whether or not order is important for each scenario and why.

1. How many 3-letter words can Brenda write using the letters E,T,A\begin{align*}E, T, A\end{align*} without repeating any letters?
2. Five different cars entered the race, each painted one of the following colors: red, orange, blue, white, purple? In how many different ways can the cars finish the race?
3. At the breakfast buffet you can take any three of the following: eggs, pancakes, potatoes, cereal, waffles. How many different 3-item breakfasts can you get?

II. Count all Permutations of n\begin{align*}n\end{align*} Objects or Events

Once you decide that the order does matter, you know that you are working with a permutation. It is helpful to know how to calculate permutations.

Let’s look at an example to see how to do this.

Example

The softball coach needs to determine how many different batting lineups she can make out of her first three batters, Able, Baker, and Chan. How many different batting orders are there?

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 batters, Able, Baker, or Chan.

Choice 1×Choice 2×Choice 3=Possible Number of Choices\begin{align*}\text{Choice} \ 1 \times \text{Choice} \ 2 \times \text{Choice} \ 3 = \text{Possible Number of Choices}\end{align*}

3 is the number of choices in Choice 1 because you have 3 batters to choose from.

2 is the number of choices in Choice 2 because you selected one batter for Choice 1 leaving 2 choices.

1 is the number of choices in Choice 3 because that is all that is left.

3×2×1=6\begin{align*}3 \times 2 \times 1 = 6\end{align*}

There are six possible choices.

Using the Counting Principle you can multiply the three choices together to get the total number of choices, or permutations, as 6. Here are the 6 different batting lineups.

Able-Baker-ChanBaker-Able-ChanChan-Able-BakerAble-Chan-BakerBaker-Chan-AbleChan-Baker-Able\begin{align*}\text{Able-Baker-Chan} \quad \quad \text{Baker-Able-Chan} \quad \quad \text{Chan-Able-Baker}\!\\ \text{Able-Chan-Baker} \quad \quad \text{Baker-Chan-Able} \quad \quad \text{Chan-Baker-Able}\end{align*}

Notice that order is important here. Each of the 6 choices, or permutations, is a unique and different batting order. For example, Able-Baker-Chan is not the same lineup as Able-Chan-Baker.

What happens when you increase the number of players in your lineups by adding Davis? How many different lineups are there now? Starting over, you can now see that there are 4 choices for the first batter, followed by 3 choices, 2 choices, and 1 choice.

4×3×2×1=24 possible choices\begin{align*}4 \times 3 \times 2 \times 1 = 24 \ \text{possible choices}\end{align*}

Let’s look at another example.

Example

How many different arrangements of the letters A,B,C,D\begin{align*}A, B, C, D\end{align*}, and E\begin{align*}E\end{align*} can you make without repeating any of the letters?

This is very much like the Able, Baker, Chan, Davis problem only it adds a fifth item. Notice how this extra item increases the total by a huge amount. In fact, there are exactly 5 times as many permutations of 5 items than there were of 4 items above.

54321=120choice 1  choice 2 choice 3 choice 4 choice 5 total choices\begin{align*}& \quad \boxed{5} \qquad \cdot \qquad \boxed{4} \qquad \cdot \qquad \boxed{3} \qquad \cdot \qquad \boxed{2} \qquad \cdot \qquad \boxed{1} \qquad = \qquad \boxed{120}\\ & \text{choice} \ 1 \qquad \ \ \text{choice} \ 2 \qquad \ \text{choice} \ 3 \qquad \ \text{choice} \ 4 \qquad \ \text{choice} \ 5 \qquad \ \text{total choices}\end{align*}

You can see how the product of all of the choices was figured out in this problem.

III. Count Permutations of n\begin{align*}n\end{align*} Objects Taken r\begin{align*}r\end{align*} at a Time

In the last section, you learned how to count permutations of a certain amount of choices where all of the choices were used each time. For example, when you completed the batter line up of three batters in order, you could count the permutations like this.

3×2×1=6 possible permutations\begin{align*}3 \times 2 \times 1 = 6 \ \text{possible permutations}\end{align*}

What happens if you were to have four players, but you only wanted to put three in the line up?

This changes the way that we count permutations.

To accomplish this task, you start counting at 4 and then find the product of the next two values as well.

4×3×2=24 permutations\begin{align*}4 \times 3 \times 2 = 24 \ \text{permutations}\end{align*}

Notice that we don’t include the 1 because there are four batters taken three at a time. Since we are only using 3 of the 4 options, we only multiply the first 3 of the 4 counts.

Let’s look at another example.

Example

What will happen if Elvis joins the team? Now, with 5 players to choose from, how many different 3-player batting orders are there?

You still have 3 different choices to make, but this time you have 5 different players for your first choice, 4 players for your second choice, and 3 players for your third choice.

5×4×3=60 options\begin{align*}5 \times 4 \times 3 = 60 \ \text{options}\end{align*}

You can use this method to find any permutation count, no matter how many options there are.

Example

Taken 4 at a time, how many different arrangements of the letters A,B,C,D,E\begin{align*}A, B, C, D, E\end{align*}, and F\begin{align*}F\end{align*} are there?

First, notice that there are six possible letter choices to work with. You want to take them four at a time.

6×5×4×3=360 possible options\begin{align*}6 \times 5 \times 4 \times 3 = 360 \ \text{possible options}\end{align*}

Notice that you don’t have to see all 360 options to know that your answer is accurate! If you followed the method of counting permutations, then your answer is correct.

IV. Evaluate Permutations Using Permutation Notation

In the last section, you figured out permutations by arranging numbers. Sometimes you used boxes to organize the numbers and sometimes you just wrote out the multiplication problem. We can use permutation notation to help us know when we are working with a permutation.

What is permutation notation?

Permutation notation involves something called a factorial. A factorial is a way of writing a number to show that we are going to be looking for the product of a series of numbers.

The symbol for a factorial is an exclamation sign.

Here are some examples of factorials.

8!11!29!2!=8factorial=11factorial=29factorial=2factorial, and so on\begin{align*}8! &= 8-\text{factorial}\\ 11! &= 11-\text{factorial}\\ 29! &= 29-\text{factorial}\\ 2! &= 2-\text{factorial, and so on}\end{align*}

To compute factorials, simply rewrite the number before the exclamation point and all of the whole numbers that come before it.

4!7!2!11!=4321=7654321=21=1110987654321\begin{align*}4! &= 4 \cdot 3 \cdot 2 \cdot 1\\ 7! &= 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\\ 2! &= 2 \cdot 1\\ 11! & = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\end{align*}

How do you calculate the values of factorial numbers? To find out, simply multiply.

4!5!7!=4321=24=54321=120=7654321=5040\begin{align*}4! &= 4 \cdot 3 \cdot 2 \cdot 1 = 24\\ 5! &= 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\\ 7! &= 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040\end{align*}

Yes. That is all that you need to know. As long as you remember how to find the product of a factorial, you will always know a short cut for permutations!

Now that you’ve learned how to work with factorials, you are ready see how to use them to calculate permutations. Suppose you have 5 letters – A,B,C,D\begin{align*}A, B, C, D\end{align*}, and E\begin{align*}E\end{align*}. You want to know how many permutations there are if you take 3 letters at a time and find all arrangements of them. In permutation notation you write this as:

5P3\begin{align*}{_5}P_3 \Longleftarrow \end{align*} 5 items taken 3 at a time

In general, permutations are written as:

nPrn\begin{align*}{_n}P_r \Longleftarrow n\end{align*} items taken r\begin{align*}r\end{align*} at a time

To compute nPr\begin{align*}{_n}P_r\end{align*} you write:

nPr=n!(nr)!= total items! (total items-items taken at a time)!\begin{align*}{_n}P_r=\frac{n!}{(n-r)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\end{align*}

To compute 5P3\begin{align*}{_5}P_3\end{align*} just fill in the numbers:

5P35P3=5!(53)!= total items! (total items-items taken at a time)!=5(4)(3)(2)(1)2(1)=1202=60\begin{align*}{_5}P_3 &= \frac{5!}{(5-3)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_5}P_3 &= \frac{5(4)(3)(2)(1)}{2(1)} = \frac{120}{2} = 60\end{align*}

There are 60 possible permutations with this example.

You can use this formula anytime that you are looking to figure out permutations!

## Real–Life Example Completed

The Order of Entertainment

Here is the original problem once again. Reread it and then work through figuring out the permutation.

Greg and Joanna are in charge of creating an order for the Talent Show. There are 6 sixth graders, 10 seventh graders and 6 eighth graders who have entered the contest.

The order that the students perform in makes a difference, and the sixth graders will all perform together. Then the seventh graders will perform and finally the eighth graders will perform.

Greg and Joanna begin with the sixth graders. Since there are six sixth graders who are performing and order does make a difference, how many different arrangements of the order are possible?

“I don’t have a clue how to figure this out,” Joanna says to Greg.

“I think I do, let me get a piece of paper.”

Greg takes out a piece of paper and writes this on it.

6×5×4×3×2×1=the number of possible arrangements\begin{align*}6 \times 5 \times 4 \times 3 \times 2 \times 1 = the \ number \ of \ possible \ arrangements\end{align*}

“You see, it makes a difference, so we can use a factorial,” Greg explains. “Now we will know how many possible arrangements of sixth graders there are.”

6×5×4×3×2×1=720\begin{align*}6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\end{align*}

There are 720 possible combinations. Greg and Joanna probably need to narrow this down a little further because that is a lot of possible arrangements. They decide to make singers one category. That will help them with possible combinations.

## Vocabulary

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

Permutation
a combination where the order matters.
Permutation Notation
using the factorial symbol to show a permutation
Factorial
a short-cut for permutations. An exclamation point next to a number is the symbol for factorial. It means "Count down from this number to 1, then multiply each of those counting numbers together into a total."

## Time to Practice

Directions: Decide whether or not order matters for each of the following scenarios. Briefly explain your reasoning.

1. Doug is going to use the following 5 letters to create his new 3-letter computer password: B, F, G, L, and T. How many different passwords can he create if he doesn’t repeat any letters?

2. Violin players in the orchestra include Jerry, Kerry, Barry, Mary, Sherry, Harry, Terri, and Perry. How many different 3-person trios can you make?

3. The 3 different numbers for Arun’s lock are 14, 35, and 20. How many different combinations must Arun try before he’ll be sure he can open his lock?

4. Mr. Chen has decided that he’s going to give Nikki, Mickey, and Hickey awards for the essay contest. What he doesn’t know is who will get 1st prize, 2nd prize, and 3rd prize. How many different ways can Mr. Chen give out the prizes?

5. Five candidates are running for 2 student senate seats: Bo, Jo, Mo, Zo, and Ro. How many different pairs of senators can there be?

6. Five skaters are competing in the County Championship Finals: Miller, Diller, Hiller, Giller, and Stiller. How many different ways can they finish first, second, third, fourth, and fifth?

Directions: Find the number of permutations for each problem.

7. How many 3-letter sequences can Brenda write using the letters A, B, and C without repeating any of the letters?

8. How many 4-digit numbers can Brenda write using the digits 1, 3, 5, 7 without repeating any of the digits?

9. Doug, Eileen, Francesca, and Garth all entered the swimming race. In how many different orders can the four racers finish?

10. Miguel is serving soup, salad, pasta, and fish for dinner. In how many different orders can he serve the four dishes?

11. Mike has 4 different playing cards: Ace, King, Jack, and Ten. How many different 4-card arrangements can he make?

12. Marlena strung 5 charms on a bracelet–a star, a fish, a diamond, a moon, and a baby shoe. Into how many different orders can she arrange the 5 charms?

13. Michelle forgot her 6-letter computer password. She knows she used the letters H, I, J, K, L, and M in the password and that she didn’t repeat any of the letters. How many different passwords must she try before she is sure to hit the correct one?

14. Seven skaters are competing in the County Championship Finals. In how many different orders can they finish first, second, third, fourth, fifth, sixth, and seventh?

Directions: Count all of the permutations.

15. Three marbles–red, blue, yellow,–are in a jar. In how many different orders can you pull two of the marbles out of the jar (without replacing either of the marbles in the jar)?

16. A green marble is added to the jar above, giving red, blue, yellow and green marbles. In how many different orders can you pull three of the marbles out of the jar (without replacing any of the marbles in the jar)?

17. In a jar with 4 marbles–red, blue, yellow, and green–how many different orders will you have if you pull just 2 marbles from the jar (without replacing either of the marbles in the jar)?

18. In a jar with 5 marbles–red, blue, yellow, green, and white–how many different orders are possible if you pull 3 marbles from the jar?

19. How many 4-digit arrangements can you make of the digits 1, 2, 3, 4, 5 if you don’t repeat any digit?

20. A TV channel has 6 different 1-hour shows to fill 3 hours of time for Thursday night. How many different program lineups can the channel present?

21. Seven ski racers compete in the finals of the slalom event. In how many different orders can the top 3 skiers finish?

22. Seven ski racers compete in the finals of the downhill event. In how many different orders can the top 4 skiers finish?

Directions: Solve each factorial.

23. 5!

24. 3!

25. 6!

26. 4!

Directions: Use permutation notation and the formula to find each permutation.

27. 4P2\begin{align*}{_4}P_2\end{align*}

28. 4P3\begin{align*}{_4}P_3\end{align*}

29. 5P4\begin{align*}{_5}P_4\end{align*}

30. 6P3\begin{align*}{_6}P_3\end{align*}

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