Have you ever tried to arrange a group of children?

Kelly is working as a camp counselor. She has to arrange a group of 8 children into the lake three at a time. The order of the children doesn't matter, but the number of children that she can take swimming at one time does matter.

How many different combinations of children are there?

Do you know how to figure this out?

**This Concept is about counting and evaluating permutations. By the end of it, you will be able to help Kelly with her problem.**

### Guidance

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 a situation to see how to do this.

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.**

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

3 is the Choice 1 because you have 3 batters to choose from.

2 is Choice 2 because you selected one batter for Choice 1 leaving 2 choices.

1 is Choice 3 because that is all that is left.

**Here is the answer:**

\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.

\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.

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

Let’s look at another situation.

How many different arrangements of the letters \begin{align*}A, B, C, D\end{align*}, and \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.

\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.

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

\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.**

\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. We will still have the correct number of permutations if we leave out the 1.*

Let’s look at another one.

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

**Again, draw a box diagram to show that 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.**

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

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

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

\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 accurate.*

You just 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 factorials.**

\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.**

\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*}

**What are the values of these factorial numbers? To find out, simply multiply.**

\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*}

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 – \begin{align*}A, B, C, D\end{align*}, and \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:

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

In general, permutations are written as:

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

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

\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 \begin{align*}{_5}P_3\end{align*} just fill in the numbers:

\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.**

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

Now try a few of these on your own.

#### Example A

\begin{align*}{_3}P_2\end{align*}

**Solution: \begin{align*}6\end{align*} possible combinations**

#### Example B

\begin{align*}{_5}P_4\end{align*}

**Solution: \begin{align*}120\end{align*} possible combinations**

#### Example C

\begin{align*}{_9}P_2\end{align*}

**Solution: \begin{align*}72\end{align*} possible combinations**

Here is the original problem once again.

Kelly is working as a camp counselor. She has to arrange a group of 8 children into the lake three at a time. The order of the children doesn't matter, but the number of children that she can take swimming at one time does matter.

How many different combinations of children are there?

To figure this out, we can use permutation notation.

We have eight children taken swimming three at a time.

\begin{align*}{_8}P_3\end{align*}

To compute \begin{align*}{_8}P_3\end{align*} just fill in the numbers:

\begin{align*}{_8}P_3 &= \frac{8!}{(8-3)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_8}P_3 &= \frac{8(7)(6)(5)(4)(3)(2)(1)}{5(4)(3)(2)(1)} = \frac{10080}{120} = 84\end{align*}

**There are 84 possible combinations of children.**

### Vocabulary

- 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 permutation.

### Guided Practice

Here is one for you to try on your own.

\begin{align*}{_9}P_5\end{align*}

**Answer**

Notice that this problem is written in permutation notation. We have figure it out using this notation.

To compute \begin{align*}{_9}P_5\end{align*} just fill in the numbers:

\begin{align*}{_9}P_5 &= \frac{9!}{(9-5)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_9}P_5 &= \frac{9(8)(7)(6)(5)(4)(3)(2)(1)}{(4)(3)(2)(1)} = \frac{362880}{24} =15,120 \end{align*}

**There are 15,120 possible combinations.**

### Video Review

- This is a James Sousa video on permutations.

### Practice

Directions: Count all of the permutations.

1. 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)?

2. 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)?

3. 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)?

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

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

6. 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?

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

8. 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.

9. 5!

10. 3!

11. 6!

12. 4!

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

13. \begin{align*}{_4}P_2\end{align*}

14. \begin{align*}{_4}P_3\end{align*}

15. \begin{align*}{_5}P_4\end{align*}

16. \begin{align*}{_6}P_3\end{align*}