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Permutation Problems

Using the nPr function found in the Math menu under PRB  on the TI calculator

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Evaluate Permutations Using Permutation Notation
License: CC BY-NC 3.0

Donna works at the library. She is arranging the new science fiction books that have come in. She only has room for six of the 20 new books on her shelf. How many different arrangements can be made when six books are chosen from a group of 20 books to be placed on a shelf?

In this concept, you will learn to evaluate permutations by using permutation notation.

Permutations

The Counting Principle states that if event can be chosen in \begin{align*}p\end{align*}p different ways and an independent event is chosen in \begin{align*}q\end{align*}q different ways, the probability of the two events occurring in \begin{align*}p \times q\end{align*}p×q. Basically, this tells you how many ways you can arrange items. 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.

How many ways can you arrange the letters in the word MATH?

First, look at the problem.

You have 4 letters in the word and you are going to choose one letter at a time. When you choose the first letter, you have 4 possibilities (M, A, T, or H), your second choice will be 3 possibilities, third choice will be 2 possibilities (because there are only 2 letters left to choose from), and last choice only one possibility.

Next, calculate the number of choices. 

\begin{align*}& P(\text{choice 1, choice 2, choice 3, choice 4)} = 4 \times 3 \times 2 \times 1 \\ & P(\text{choice 1, choice 2, choice 3, choice 4)} = 24\end{align*}P(choice 1, choice 2, choice 3, choice 4)=4×3×2×1P(choice 1, choice 2, choice 3, choice 4)=24

The answer is 24.

There are 24 different ways to arrange the letters in the word MATH.

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 at how factorials are used.

\begin{align*}8 \ \text{factorial} = 8!= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\end{align*}8 factorial=8!=8×7×6×5×4×3×2×1

\begin{align*}11 \ \text{factorial}= 11!= 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\end{align*}11 factorial=11!=11×10×9×8×7×6×5×4×3×2×1

\begin{align*}4 \ \text{factorial} = 4! = 4 \times 3 \times 2 \times 1\end{align*}4 factorial=4!=4×3×2×1

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

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

\begin{align*}\begin{array}{rcl} 4 \ \text{factorial} = 4! &=& 4 \times 3 \times 2 \times 1 \\ &=& 24\\ 6 \ \text{factorial} = 6! &=& 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ & =& 720\\ 8 \ \text{factorial} = 8! &=& 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ &=& 40,320 \end{array}\end{align*}4 factorial=4!6 factorial=6!8 factorial=8!======4×3×2×1246×5×4×3×2×17208×7×6×5×4×3×2×140,320

You can also use a graphing calculator to find factorials. For large numbers, especially, the calculator can save you time.

If you push the \begin{align*}m\end{align*}m button and on the top of the screen you will see PROB. #4 under the PROB menu is factorial.

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Let’s try an example.

Find 10! using your calculator.

License: CC BY-NC 3.0

The answer is 3628800. Notice the key press history to tell you what was pressed to get 10!

You can use factorials to calculate permutations. Look at the formula for finding permutations.

In general, permutations are written as:

\begin{align*}{\color{red}_n}P{\color{blue}_r} = {\color{red}n} \ \text{items taken} \ {\color{blue}r} \ \text{at a time}\end{align*}nPr=n items taken r at a time

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

\begin{align*}_nP _r = \frac{n !}{(n - r)!}\end{align*}nPr=n!(nr)!

Let’s look at an example.

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

First, order matters in this problem, so you need to find the number of permutations there are in 6 items taken 4 at a time.

First, in permutation notation write the following.

\begin{align*}{\color{red}_6}P{\color{blue}_4} = \end{align*}6P4= 6 items taken 4 at a time

Next, calculate \begin{align*}_6P_4\end{align*}6P4.

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _6P _4 &=& \frac{6 !}{(6 - 4)!} \\ \\ _6P _4 &=& \frac{6 !}{2 !} \\ \\ _6P _4 &=& \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} \\ \\ _6P _4 &=& 6 \times 5 \times 4 \times 3 \\ \\ _6P _4 &=& 360 \end{array}\end{align*}nPr6P46P46P46P46P4======n!(nr)!6!(64)!6!2!6×5×4×3×2×12×16×5×4×3360

Notice that it is the product of the values in descending order that tells you how many permutations are possible. You can use this method to solve any number of permutations.

You can also use the graphing calculator to find \begin{align*}_nP_r\end{align*}nPr. If you push the m button and on the top of the screen you will see PROB. #2 under the PROB menu is \begin{align*}_nP_r\end{align*}nPr.

License: CC BY-NC 3.0

To find \begin{align*}_8P_3\end{align*}8P3 using the calculator, you would get the following. Notice the key press history to tell you what to press to do \begin{align*}_8P_3\end{align*}8P3.

License: CC BY-NC 3.0

Examples

Example 1

Earlier, you were given a problem about Donna the librarian.

Donna is trying to choose six books of her 20 new books to put on the shelf. Remember that order is important and repetition is not allowed, so permutations must be used.

Donna has 20 books to choose from, therefore \begin{align*}n = 20\end{align*}n=20. She will choose 6 books, therefore \begin{align*}r = 6\end{align*}r=6.

First, fill in the formula for \begin{align*} _nP_r\end{align*}nPr where \begin{align*}n = 20\end{align*}n=20 and \begin{align*}r = 6\end{align*}r=6.

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _{20}P _6 &=& \frac{20 !}{(20 - 6)!} \end{array}\end{align*}nPr20P6==n!(nr)!20!(206)!

Next, compute \begin{align*}_{20}P_6\end{align*}20P6.

\begin{align*}\begin{array}{rcl} _{20}P _6 & = & \frac{20 !}{(20 - 6)!} \\ \\ _{20}P _6 & = & \frac{20 !}{14 !} \\ \\ _{20}P _6 & = & \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13\times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{14 \times 13\times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ \\ _{20}P _6 & = & 20 \times 19 \times 18 \times 17 \times 16 \times 15 \\ \\ _{20}P _6 & = & 27907200 \end{array}\end{align*}20P620P620P620P620P6=====20!(206)!20!14!20×19×18×17×16×15×14×13×12×11×10×9×8×7×6×5×4×3×2×114×13×12×11×10×9×8×7×6×5×4×3×2×120×19×18×17×16×1527907200

The answer is 27907200.

There are 27,907,200 different ways Donna can choose six books of the 20 to display on the shelf.

Example 2

Find \begin{align*}_7P_3\end{align*}.

First, fill in the formula for \begin{align*}_nP_r\end{align*} where \begin{align*}n = 7\end{align*} and \begin{align*}r = 3\end{align*}

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _7P _3 &=& \frac{7 !}{(7 - 3)!} \end{array}\end{align*}

Next, compute \begin{align*}_7P_3\end{align*}.

\begin{align*}\begin{array}{rcl} _7P _3 &=& \frac{7 !}{(7 - 3)!} \\ \\ _7P _3 &=& \frac{7 !}{4 !} \\ \\ _7P _3 &=& \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \\ \\ _7P _3 &=& 7 \times 6 \times 5 \\ \\ _7P _3 &=& 210 \end{array}\end{align*}

The answer is 210.

Example 3

Find \begin{align*}_4P_3\end{align*}.

First, fill in the formula for \begin{align*}_nP_r\end{align*} where \begin{align*}n = 4\end{align*} and \begin{align*}r = 3\end{align*}

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _4P _3 &=& \frac{4 !}{(4 - 3)!} \end{array}\end{align*}

Next, compute \begin{align*}_4P_3\end{align*}.

\begin{align*}\begin{array}{rcl} _4P _3 &=& \frac{4 !}{(4 - 3)!} \\ \\ _4P _3 &=& \frac{4 !}{1 !} \\ \\ _4P _3 &=& \frac{4 \times 3 \times 2 \times 1}{1} \\ \\ _4P _3 &=& 4 \times 3 \times 2 \times 1 \\ \\ _4P _3 &=& 24 \end{array}\end{align*}

The answer is 24.

Example 4

Find \begin{align*}_{12}P_2\end{align*}.

First, fill in the formula for \begin{align*}_nP_r\end{align*} where \begin{align*}n = 12\end{align*} and \begin{align*}r = 2\end{align*}

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _{12}P _2 &=& \frac{12 !}{(12 - 2)!} \end{array}\end{align*}

Next, compute \begin{align*} _{12}P_2\end{align*}.

\begin{align*}\begin{array}{rcl} _{12}P _2 &=& \frac{12 !}{(12 - 2)!} \\ \\ _{12}P _2 &=& \frac{12 !}{10 !} \\ \\ _{12}P _2 &=& \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ \\ _{12}P _2 &=& 12 \times 11 \\ \\ _{12}P _2 &=& 132 \end{array}\end{align*}

The answer is 132.

Example 5

Find \begin{align*}_8P_6\end{align*}.

First, fill in the formula for \begin{align*}_nP_r\end{align*} where \begin{align*}n = 8\end{align*} and \begin{align*}r = 6\end{align*}

\begin{align*}\begin{array}{rcl} _nP _r &=& \frac{n !}{(n - r)!} \\ \\ _8P _6 &=& \frac{8 !}{(8 - 6)!} \end{array}\end{align*}

Next, compute \begin{align*}_8P_6\end{align*}.

\begin{align*}\begin{array}{rcl} _8P _6 &=& \frac{8 !}{(8 - 6)!} \\ \\ _8P _6 &=& \frac{8 !}{2 !} \\ \\ _8P _6 &=& \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{ 2 \times 1} \\ \\ _8P _6 &=& 8 \times 7 \times 6 \times 5 \times 4 \times 3 \\ \\ _8P _6 &=& 20160 \end{array}\end{align*}

The answer is 20,160.

Review

Find each permutation.

1. Find \begin{align*}_7P_2\end{align*}

2. Find \begin{align*}_6P_3\end{align*}

3. Find \begin{align*}_5P_4\end{align*}

4. Find \begin{align*}_5P_5\end{align*}

5. Find \begin{align*}_9P_3\end{align*}

6. Find \begin{align*}_9P_7\end{align*}

7. Find \begin{align*}_{11}P_3\end{align*}

8. Find \begin{align*}_{12}P_3\end{align*}

9. Find \begin{align*}_6P_2\end{align*}

10. Find \begin{align*}_{14}P_3\end{align*}

11. Find \begin{align*}_{15}P_3\end{align*}

12. Find \begin{align*}_{11}P_4\end{align*}

13. Find \begin{align*}_{16}P_2\end{align*}

Use permutations to solve each problem.

14. 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?

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

Review (Answers)

To see the Review answers, open this PDF file and look for section 11.4. 

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