Do you know how to calculate a permutation in permutation notation? Take a look at this dilemma.

Find \begin{align*} _{8}P _{2}\end{align*}

**
To figure out this permutation, you will need to understand permutation notation. Pay attention and you will learn all that you need to know in this Concept.
**

### Guidance

*
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.

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.
**

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. Take a look.

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

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

@$$\begin{align*}4! &= 4 \cdot 3 \cdot 2 \cdot 1 = 24\\ 5! &= 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\\ 8! &= 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 40,320\\ 11! &= 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 39,916,800\end{align*}@$$

**
We can use factorials to calculate permutations.
**

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.

@$\begin{align*}{\color{red}_{6}}P {\color{blue}_{4}} \ \Longleftarrow \end{align*}@$ 6 items taken 4 at a time

In general, permutations are written as:

@$\begin{align*}{\color{red}_{n}}P {\color{blue}_{r}} \ \Longleftarrow {\color{red}n}\end{align*}@$ items taken @$\begin{align*}{\color{blue}r}\end{align*}@$ at a time

To compute @$\begin{align*}{_n}P{_r}\end{align*}@$ you write:

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

To compute @$\begin{align*}_{6}P _{4}\end{align*}@$ just fill in the numbers:

@$$\begin{align*} {\color{red}_{6}}P {\color{blue}_{4}}&= \frac{{\color{red}6}!}{({\color{red}6}-{\color{blue}4})!}\end{align*}@$$

@$\begin{align*}\frac{720}{2}\end{align*}@$

@$\begin{align*}360\end{align*}@$

**
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.
**

Find each permutation.

#### Example A

Find @$\begin{align*} _{4}P _{3}\end{align*}@$

**
Solution:
@$\begin{align*}24\end{align*}@$
options
**

#### Example B

Find @$\begin{align*} _{12}P _{2}\end{align*}@$

**
Solution:
@$\begin{align*}132\end{align*}@$
options
**

#### Example C

Find @$\begin{align*} _{8}P _{6}\end{align*}@$

**
Solution:
@$\begin{align*}20,160\end{align*}@$
options
**

Now let's go back to the dilemma from the beginning of the Concept.

Find @$\begin{align*} _{8}P _{2}\end{align*}@$

This is the same as saying 8 options taken 2 at a time.

We can multiply to solve the permutation.

@$\begin{align*}8 \times 7 = 56\end{align*}@$

**
There are 56 options.
**

### Guided Practice

Here is one for you to try on your own.

Find @$\begin{align*} _{7}P _{3}\end{align*}@$

**
Solution
**

**
Step 1
:
**
Understand what
@$\begin{align*} _{7}P _{3}\end{align*}@$
means.

@$\begin{align*} {\color{red}_{7}}P {\color{blue}_{3}} \ \Longleftarrow \end{align*}@$ 7 items taken 3 at a time

**
Step 2
:
**
Set up the problem.

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

**
Step 3
:
**
Fill in the numbers and simplify.

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

**
There are 210 possible permutations.
**

### Video Review

### Explore More

Directions: Find each permutation.

- Find @$\begin{align*} _{7}P _{2}\end{align*}@$
- Find @$\begin{align*} _{6}P _{3}\end{align*}@$
- Find @$\begin{align*} _{5}P _{4}\end{align*}@$
- Find @$\begin{align*} _{5}P _{5}\end{align*}@$
- Find @$\begin{align*} _{9}P _{3}\end{align*}@$
- Find @$\begin{align*} _{9}P _{7}\end{align*}@$
- Find @$\begin{align*} _{11}P _{3}\end{align*}@$
- Find @$\begin{align*} _{12}P _{3}\end{align*}@$
- Find @$\begin{align*} _{6}P _{2}\end{align*}@$
- Find @$\begin{align*} _{14}P _{3}\end{align*}@$
- Find @$\begin{align*} _{15}P _{3}\end{align*}@$
- Find @$\begin{align*} _{11}P _{4}\end{align*}@$
- Find @$\begin{align*} _{16}P _{2}\end{align*}@$

Directions: Use permutations to solve each problem.

- 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?
- One of the charms in Mia’s bracelet in problem 6 above fell off. How many fewer arrangements are there now?