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# Permutations

## The number of arrangements when order matters

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Define and Apply Permutations and Factorials

Your job as county fair assistant is to arrange the sheep ribbons on the bulletin board overhanging the sheep pens. You have one Best in Show ribbon, one 1st Place ribbon, one 2nd Place ribbon, and one 3rd Place ribbon to display. How many ways can you arrange the ribbons?

### Permutations and Factorials

The number of permutations of objects is the number of possible arrangements of the objects. Consider the following question: How many ways can seven DVD’s be arranged on a shelf? This is an example of a permutation. We will use the Fundamental Counting Principle without repetition to determine the permutations.

Let's solve the following problem using permutations.

How many ways can 5 students sit in a row?

If we consider the students sitting in one of five seats, then we have 5 students to choose from for the first seat, four remaining to choose from for the second seat and so on until all the seats are filled.

5×4×3×2×1=120\begin{align*}\underline5\times\underline4\times\underline3\times\underline2\times\underline1=120\end{align*} So there are 120 ways to seat the students.

The way we just wrote out 5×4×3×2×1\begin{align*}\underline5\times\underline4\times\underline3\times\underline2\times\underline1\end{align*} can also be expressed as a factorial. A factorial is the product of a number with each number less than itself. We use the notation, 5!\begin{align*}5!\end{align*}, which is read as “five factorial” to represent the expression 5×4×3×2×1\begin{align*}\underline5\times\underline4\times\underline3\times\underline2\times\underline1\end{align*}. It is important to note that 0!=1!=1\begin{align*}0!=1!=1\end{align*}. Students are often perplexed that both zero and one factorial are equal to one but think back to the context for illustration. If you want to arrange zero items, how many ways can you do it? If you want to arrange one item, how many ways can you do it? There is only one way to “arrange” zero or one item.

To evaluate a factorial on the TI-83 or 84 graphing calculator, type in the number, then press MATH\begin{align*}\rightarrow\end{align*}NUM, 4!\begin{align*}4!\end{align*}. Press ENTER to evaluate.

Now, let's solve the following problems using factorials.

1. Evaluate 10!6!\begin{align*}\frac{10!}{6!}\end{align*}.

We should expand the numerator and the denominator to see which common factors the numerator and denominator that we can cancel out to simplify the expression.

10×9×8×7×6×5×4×3×2×16×5×4×3×2×1=10×9×8×7×6!6!=10×9×8×7=5040\begin{align*}\frac{10\times9\times8\times7\times{\color{red}6\times5\times4\times3\times2\times1}}{{\color{red}6\times5\times4\times3\times2\times1}}=\frac{10\times9\times8\times7\times{\color{red}\cancel{6!}}}{{\color{red}\cancel{6!}}}=10\times9\times8\times7=5040\end{align*}

1. On a shelf there are 6 different math books, 4 different science books and 8 novels. How many ways can the books be arranged if the groupings are maintained (meaning all the math books are together, the science books are together and the novels are together).

There are 6 math books so if we think of filling 6 slots with the six books, then we start with 6 books for the first slot, then 5, then 4, etc: 6×5×4×3×2×1=720\begin{align*}{\underline{{\color{red}6}}\times\underline{{\color{red}5}}\times\underline{{\color{red}4}}\times\underline{{\color{red}3}}\times\underline{{\color{red}2}}\times\underline{{\color{red}1}}}={\color{red}720}\end{align*} ways.

There are 4 science books so we can arrange them in 4×3×2×1=24\begin{align*}\underline{{\color{blue}4}}\times\underline{{\color{blue}3}}\times\underline{{\color{blue}2}}\times\underline{{\color{blue}1}}={\color{blue}24}\end{align*} ways.

There are 8 novels so we can arrange them in 8×7×6×5×4×3×2×1=40,320\begin{align*}\underline{{\color{green}8}}\times\underline{{\color{green}7}}\times\underline{{\color{green}6}}\times\underline{{\color{green}5}}\times\underline{{\color{green}4}}\times\underline{{\color{green}3}}\times\underline{{\color{green}2}}\times\underline{{\color{green}1}}={\color{green}40,320}\end{align*} ways.

Now, if each type of book can be arranged in so many ways and there are three types of books which can be displayed in 3×2×1=6\begin{align*}\underline{3}\times\underline{2}\times\underline{1}=6\end{align*} ways, then there are:

720×24×40320×6=4,180,377,600\begin{align*}{\color{red}720}\times{\color{blue}24}\times{\color{green}40320}\times6=4,180,377,600\end{align*} total ways to arrange the books.

### Examples

#### Example 1

Earlier, you were asked to find the number of ways you can arrange the ribbons.

If we consider the ribbons sitting in one of four spots, then we have four ribbons to choose from for the first spot, three remaining to choose from for the second spot and so on until all the spots are filled.

4×3×2×1=24\begin{align*}\underline4\times\underline3\times\underline2\times\underline1=24\end{align*}

Therefore, there are 24 ways to arrange the ribbons.

For Examples 2 & 3, evaluate the expressions with factorials.

#### Example 2

12!9!\begin{align*}\frac{12!}{9!}\end{align*}

12×11×10×9!9!=1320\begin{align*}\frac{12\times11\times10\times{\color{red}\cancel{9!}}}{{\color{red}\cancel{9!}}}=1320\end{align*}

#### Example 3

4×8!3!5!\begin{align*}\frac{4\times8!}{3!5!}\end{align*}

4×8×7×6×5!3×2×1×5!=4×8×7×63×2×1=224\begin{align*}\frac{4\times8\times7\times6\times{\color{red}\cancel{5!}}}{3\times2\times1\times{\color{red}\cancel{5!}}}=\frac{4\times8\times7\times{\color{red}\cancel{6}}}{{\color{red}\cancel{3}\times\cancel{2}}\times1}=224\end{align*}

#### Example 4

How many ways can nine children line up?

9!=362,880\begin{align*}9!=362,880\end{align*}

#### Example 5

How many ways can 3 cookbooks, 5 textbooks, 7 novels and 4 nonfiction books be arranged on a shelf if the groupings are maintained?

3!×5!×7!×4!×4!=2,090,188,800\begin{align*}3!\times5!\times7!\times4!\times4!=2,090,188,800\end{align*}

### Review

Evaluate the following factorial expressions.

1. 5!2!3!\begin{align*}\frac{5!}{2!3!}\end{align*}
2. 10!2!7!\begin{align*}\frac{10!}{2!7!}\end{align*}
3. 4!8!9!\begin{align*}\frac{4!8!}{9!}\end{align*}
4. 5!10!12!\begin{align*}\frac{5!10!}{12!}\end{align*}
5. How many ways can a baseball team manager arrange nine players in a lineup?
6. How many ways can the letters in the word FACTOR be arranged?
7. How many ways can 12 school buses line up?
8. How many ways can eight girls sit together in a row?
9. If the two of the eight girls in problem eight must sit together, how many ways can the 8 girls be arranged in the row such that the two girls sit together?
10. How many ways can seven diners sit around a circular table. (Hint: It is not 7!\begin{align*}7!\end{align*}, consider how a circular seating arrangement is different than a linear arrangement.)
11. How many ways can three cookbooks, four novels and two nonfiction books be arranged on a shelf if the groupings are maintained?
12. How many ways can two teachers, four male students, five female students and one administrator be arranged if the teachers must sit together, the male students must sit together and the female students must sit together?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 12.3.

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### Vocabulary Language: English

Permutation

A permutation is an arrangement of objects where order is important.