<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

Permutation Problems

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

Atoms Practice
Estimated6 minsto complete
%
Progress
Practice Permutation Problems
Practice
Progress
Estimated6 minsto complete
%
Practice Now
Permutation Problems

Let's Think About It

Veronica is a lifeguard at the community swimming pool.  She is in charge of creating an order for the 10 children who are waiting to use the diving board.  She can send them four at a time to line up.  How many different combinations can Veronica make?

In this concept, you will learn about counting permutations.

Guidance

Once you decide that 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.

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*}

Choice 1×Choice 2×Choice 3=Possible Number of Choices

3 is 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.

The answer is \begin{align*}3 \times 2 \times 1 = 6\end{align*}3×2×1=6.  There are six possible orders.

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*}

Able-Baker-ChanBaker-Able-ChanChan-Able-BakerAble-Chan-BakerBaker-Chan-AbleChan-Baker-Able

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*}

4×3×2×1=24 possible choices

Let’s look at another example.

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

54321=120choice 1  choice 2 choice 3 choice 4 choice 5 total choices

You learned how to count permutations 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*}

3×2×1=6 possible permutations

What happens if you have four players, but you only wanted to put three in the line up?  This changes the way that you 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*}

4×3×2=24 permutations

Notice that you don’t include the 1 because there are four batters taken three at a time. 

Let’s look at another 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?

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*}

5×4×3=60 options

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

6×5×4×3=360 possible options

You just figured out permutations by arranging numbers.  You can use permutation notation to help you know when you are working with a permutation.  Permutation notation involves something called a factorial. A factorial is a way of writing a number to show that you 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*}

8!11!29!2!=8factorial=11factorial=29factorial=2factorial, and so on

To compute factorials, 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*}

4!7!2!11!=4321=7654321=21=1110987654321

What are the values of these factorial numbers?

\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*}

4!5!7!=4321=24=54321=120=7654321=5040

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*}A,B,C,D, and \begin{align*}E\end{align*}E. 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*}5P3 5 items taken 3 at a time

In general, permutations are written as:

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

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

nPr=n!(nr)!= total items! (total items-items taken at a time)!

To compute \begin{align*}{_5}P_3\end{align*}5P3 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*}

5P35P3=5!(53)!= total items! (total items-items taken at a time)!=5(4)(3)(2)(1)2(1)=1202=60

There are 60 possible permutations.

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

Guided Practice

Calculate the following permutation:

\begin{align*}{_9}P_5\end{align*}9P5

First, substitute the values into the appropriate place in the formula:

\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)} \end{align*}

9P59P5=9!(95)!= total items! (total items-items taken at a time)!=9(8)(7)(6)(5)(4)(3)(2)(1)(4)(3)(2)(1)

Next, multiply all values together:

\begin{align*}{_9}P_5 &= \frac{9(8)(7)(6)(5)(4)(3)(2)(1)}{(4)(3)(2)(1)} = \frac{362880}{24} \end{align*}

9P5=9(8)(7)(6)(5)(4)(3)(2)(1)(4)(3)(2)(1)=36288024

Then, divide the numerator by the denominator:

\begin{align*}{_9}P_5 &= \frac{362880}{24} =15,120 \end{align*}

9P5=36288024=15,120
The answer is there are 15,120 possible combinations.

Examples

Calculate the following permutations.

Example 1

\begin{align*}{_3}P_2\end{align*}3P2

First, substitute the values into the appropriate place in the formula:\begin{align*}{_3}P_2 &= \frac{3!}{(3-2)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_3}P_2 &= \frac{(3)(2)(1)}{(1)} \end{align*}

3P23P2=3!(32)!= total items! (total items-items taken at a time)!=(3)(2)(1)(1)
Next, multiply all values together:\begin{align*}{_3}P_2 &= \frac{(3)(2)(1)}{(1)} = \frac{6}{1} \end{align*}
3P2=(3)(2)(1)(1)=61
Then, divide the numerator by the denominator:\begin{align*}{_3}P_2 &= \frac{6}{1} =6 \end{align*}
3P2=61=6
The answer is there are 6 possible combinations.

Example 2

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

First, substitute the values into the appropriate place in the formula:\begin{align*}{_5}P_4 &= \frac{5!}{(5-4)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_5}P_4 &= \frac{(5)(4)(3)(2)(1)}{(1)} \end{align*}

Next, multiply all values together:\begin{align*}{_5}P_4 &= \frac{(5)(4)(3)(2)(1)}{(1)} = \frac{120}{1} \end{align*}
Then, divide the numerator by the denominator:\begin{align*}{_5}P_4 &= \frac{120}{1} =120 \end{align*}
The answer is there are 120 possible combinations.

Example 3

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

First, substitute the values into the appropriate place in the formula:\begin{align*}{_9}P_2 &= \frac{9!}{(9-2)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_9}P_2 &= \frac{9(8)(7)(6)(5)(4)(3)(2)(1)}{(7)(6)(5)(4)(3)(2)(1)} \end{align*}

Next, multiply all values together:\begin{align*}{_9}P_2 &= \frac{9(8)(7)(6)(5)(4)(3)(2)(1)}{(7)(6)(5)(4)(3)(2)(1)} = \frac{362,880}{5,040} \end{align*}
Then, divide the numerator by the denominator:\begin{align*}{_9}P_2 &= \frac{362880}{5,040} =72\end{align*}
The answer is there are 72 possible combinations.

Follow Up

Remember Veronica's responsibility of sending four children to line up at the diving board at a time?  There are a total of 10 children.  How many different combinations  of children can Veronica make?

First, substitute the values into the appropriate place in the formula.

\begin{align*}{_10}P_4 &= \frac{10!}{(10-4)!}=\frac{\Longleftarrow \ \text{total items!}}{\Longleftarrow \ (\text{total items-items taken at a time})!}\\ {_10}P_4 &= \frac{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{(6)(5)(4)(3)(2)(1)} \end{align*}

Next, multiply all values together.\begin{align*}{_10}P_4 &= \frac{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{(6)(5)(4)(3)(2)(1)} = \frac{3,628,800}{720} \end{align*}
Then, divide the numerator by the denominator.\begin{align*}{_10}P_4 &= \frac{3,628,800}{720} =5,040 \end{align*}
The answer is there 5,040 possible combinations.

Video Review

Explore More

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?

Solve each factorial.

9. 5!

10. 3!

11. 6!

12. 4!

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*}

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.13. 

Vocabulary

n value

n value

When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing, and the r value is the number of objects chosen.
r value

r value

When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing, and the r value is the number of objects chosen.
combination

combination

Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
factorial

factorial

The factorial of a whole number n is the product of the positive integers from 1 to n. The symbol "!" denotes factorial. n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n .
n value

n value

When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing.
Permutation

Permutation

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

r value

When calculating permutations with the TI calculator, the r value is the number of objects chosen.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Permutation Problems.

Reviews

Please wait...
Please wait...

Original text