# 7.8: Probability and Permutations

**Basic**Created by: CK-12

**Practice**Probability and Permutations

Suppose there are 15 people in a meeting, and one person will be the facilitator, while another person will be the scribe. In how many different ways can the facilitator and the scribe be chosen? What formula do you think you can use to answer this question? If the facilitator and the scribe were chosen at random, could you determine the probability of any particular pair of people?

### Probability and Permutations

Congratulations! You have won a free trip to Europe. On your trip you have the opportunity to visit 6 different cities. You are responsible for planning your European vacation. How many different ways can you schedule your trip? The answer may surprise you!

This is an example of a permutation.

A **permutation** is an arrangement of objects in a specific order. Calculating a permutation involves finding the factorial of a number. A **factorial **is indicated by using an exclamation mark after the number and is the product of the counting numbers 1 through \begin{align*}n\end{align*}

\begin{align*}n!=n(n-1)(n-2)\cdot \ldots \cdot 1\end{align*}

How many ways can you visit the European cities? There are 6 choices for the first stop. Once you have visited this city, you cannot return so there are 5 choices for the second stop, and so on.

\begin{align*}\underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=720\end{align*}

There are 720 different ways to plan your European vacation!

A permutation of \begin{align*}n\end{align*}**objects** arranged \begin{align*}k\end{align*}**at a time** is expressed as \begin{align*}_n P_k\end{align*}

#### \begin{align*}_n P_k =\frac{n!}{(n-k)!}\end{align*}nPk=n!(n−k)!

The problem about the European trip from above can also be written using the formula for permutations: \begin{align*}_6P_6=\frac{6!}{(6-6)!}=\frac{6!}{0!}=6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720\end{align*}

#### Let's evaluate \begin{align*}_6P_3\end{align*}6P3 .

This equation asks, “How many ways can 6 objects be chosen 3 at a time?”

Using the formula from above, we get:

\begin{align*}_6P_3&=\frac{6!}{(6-3)!}\\
&=\frac{6\cdot 5\cdot 4\cdot 3 \cdot 2\cdot 1}{3\cdot 2\cdot 1}\end{align*}

Notice, that some of the factors can be cancelled out. It is a good idea to do this before you continue simplifying.

\begin{align*}_6P_3&=\frac{6\cdot 5\cdot 4\cdot 3 \cdot 2\cdot 1}{3\cdot 2\cdot 1}\\
&=\frac{6\cdot 5\cdot 4\cdot \cancel 3 \cdot \cancel 2\cdot \cancel 1}{\cancel 3\cdot \cancel 2\cdot \cancel 1}\\
&=6\cdot 5\cdot 4\\
&=120\end{align*}

This result makes sense logically as there are 6 ways to choose the first object, 5 ways to choose the second object, and 4 ways to choose the third object.

\begin{align*}\underline{6} \cdot \underline{5} \cdot \underline{4}=120\end{align*}

There are 120 different ways 6 objects can be chosen 3 at a time.

#### Permutations and Graphing Calculators

Most graphing calculators have the ability to calculate a permutation.

#### Let's see the steps by evaluating \begin{align*}_6P_3\end{align*}6P3 using a graphing calculator:

Type the first value of the permutation, the \begin{align*}n\end{align*}**[MATH]** button, directly below the **[ALPHA]** key. Move the cursor once to the left to see this screen:

Option #2 is the permutation option. Press **[ENTER]** and then the second value of the permutation, the value of \begin{align*}k\end{align*}**[ENTER]** to evaluate.

#### Probability with Permutations

Permutations and factorials are very useful when calculating the probability for certain events.

#### Let's find how many different arrangements can be found using the letters of the word HOSPITAL and find what the probability that the last letter is a vowel is.

There are eight ways to choose the first letter, seven ways to choose the second, and so on. The total number of arrangements is 8!= 40,320.

There are three vowels in HOSPITAL; therefore, there are three possibilities for the last letter. Once this letter is chosen, there are seven choices for the first letter, six for the second, and so on.

\begin{align*}\underline{7} \cdot \underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1} \cdot \underline{3}=15,120\end{align*}

Probability, as you learned in a previous Concept, has the formula:

\begin{align*}\text{Probability} \ (success) = \frac{number \ of \ ways \ to \ get \ success}{total \ number \ of \ possible \ outcomes}\end{align*}

There are 15,120 ways to get a vowel as the last letter; there are 40,320 total combinations.

\begin{align*}P(last \ letter \ is \ a \ vowel)=\frac{15,120}{40,320}=\frac{3}{8}\end{align*}

### Examples

#### Example 1

Earlier, you were told that there are 15 people in the meeting, and one person will be the facilitator while another person will be the scribe. In how many different ways can the facilitator and the scribe be chosen? If the facilitator and the scribe were chosen at random, could you determine the probability of any particular pair of people?

There are 15 people and we are choosing two people out of those 15. The permutation for this would be \begin{align*}_{15} P_2\end{align*}

Let's solve this permutation:

\begin{align*}_{15}P_2&=\frac{15!}{(15-2)!}
\\&=\frac{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}{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}\\
&=\frac{15\cdot 14\cdot \cancel {13} \cdot \cancel {12}\cdot \cancel{11}\cdot \cancel {10}\cdot \cancel 9 \cdot \cancel 8\cdot \cancel 7\cdot \cancel 6 \cdot \cancel 5\cdot \cancel 4\cdot \cancel 3 \cdot \cancel 2\cdot \cancel 1}{\cancel {13} \cdot \cancel {12}\cdot \cancel{11}\cdot \cancel {10}\cdot \cancel 9 \cdot \cancel 8\cdot \cancel 7\cdot \cancel 6 \cdot \cancel 5\cdot \cancel 4\cdot \cancel 3 \cdot \cancel 2\cdot \cancel 1}\\
&=15\cdot 14\\
&=210\end{align*}

There are 210 ways to choose the facilitator and the scribe.

To determine the probability of any particular pair of people, we need to know the number of ways to choose that pair of people and the number of ways to choose any two pairs of people. There is only 1 way to select a certain pair of people and we found the total possible ways to select pairs of people above. Therefore, the probability for choosing a particular pair of people is \begin{align*}\frac{1}{210}\end{align*}

#### Example 2

You have 8 different colors of nail polish. How many ways can you paint each of your 5 finger nails on one hand a different color?

There are 8 colors to paint, but you will only use 5 of them to paint your nails, which are in a particular order. This can be represented as \begin{align*}_8P_5\end{align*}

\begin{align*}_8P_5=8\cdot 7\cdot 6\cdot 5\cdot 4=6,720\end{align*}

#### Example 3

Using the information from Example 2, suppose 1 of your nail colors is purple. What is the probability that your thumb will be painted purple?

The number of ways to paint your 5 nails different colors, where the thumb is purple, is calculated as:

\begin{align*}1\cdot 7\cdot 6\cdot 5\cdot \cdot 4=840 \end{align*}

### Review

- Define permutation.

In 2 – 19, evaluate each permutation.

- 7!
- 10!
- 1!
- 5!
- 9!
- 3!
- \begin{align*}4!+4!\end{align*}
4!+4! - \begin{align*}16!-5!\end{align*}
16!−5! - \begin{align*}\frac{98!}{96!}\end{align*}
98!96! - \begin{align*}\frac{11!}{2!}\end{align*}
- \begin{align*}\frac{301!}{300!}\end{align*}
- \begin{align*}\frac{8!}{3!}\end{align*}
- \begin{align*}2!+9!\end{align*}
- \begin{align*}_{11} P_2\end{align*}
- \begin{align*}_5P_5\end{align*}
- \begin{align*}_5P_3\end{align*}
- \begin{align*}_{15}P_{10}\end{align*}
- \begin{align*}_{60}P_{59}\end{align*}
- How many ways can 14 books be organized on a shelf?
- How many ways are there to choose 10 objects, four at a time?
- How many ways are there to choose 21 objects, 13 at a time?
- A running track has eight lanes. In how many ways can 8 athletes be arranged to start a race?
- Twelve horses run a race.
- How many ways can first and second places be won?
- How many ways can all the horses finish the race?

- Six actors are waiting to audition. How many ways can the director choose the audition schedule?
- Jerry, Kerry, Larry, and Mary are waiting at a bus stop. What is the probability that Mary will get on the bus first?
- How many permutations are there of the letters in the word “HEART”?
- How many permutations are there of the letters in the word “AMAZING”?
- Suppose I am planning to get a three-scoop ice cream cone with chocolate, vanilla, and Superman. How many ice cream cones are possible? If I ask the server to “surprise me,” what is the probability that the Superman scoop will be on top?
- What is the probability that you choose two cards (without replacement) from a standard 52-card deck and both cards are jacks?
- The Super Bowl Committee has applications from 9 towns to host the next two Super Bowls. How many ways can they select the host if:
- The town cannot host a Super Bowl two consecutive years?
- The town can host a Super Bowl two consecutive years?

**Mixed Review**

- Graph the solution to the following system: \begin{align*}& 2x-3y > -9\\ & y<1\end{align*}
- Convert 24 meters/minute to feet/second.
- Solve for \begin{align*}t: |t-6| \le -14\end{align*}.
- Find the distance between 6.15 and –9.86.
- Which of the following vertices provides the minimum cost according to the equation \begin{align*}12x+20y=cost: \ (3,6),(9,0),(6,2),(0,11)\end{align*}?

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.8.

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Permutation

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

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

A favorable outcome is the outcome that you are looking for in an experiment.Probability

Probability is the chance that something will happen. It can be written as a fraction, decimal or percent.### Image Attributions

Here you'll learn how to find the number of permutations in a given situation and how to use this number in calculations of probability.

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