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# 7.7: Probability and Permutations

Difficulty Level: At Grade Created by: CK-12

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. It is the product of the counting numbers 1 through n\begin{align*}n\end{align*}.

n!=n(n1)(n2)1\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.

654321=720\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 n\begin{align*}n\end{align*} objects arranged k\begin{align*}k\end{align*} at a time is expressed as nPk\begin{align*}_n P_k\end{align*}.

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

Example 1: Evaluate 6P3\begin{align*}_6P_3\end{align*}.

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

There are 6 ways to choose the first object, 5 ways to choose the second object, and 4 ways to choose the third object.

654=120\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.

Example 1 can also be written using the formula for permutation: 6P3=6!(63)!=6!3!=654=120\begin{align*}_6P_3=\frac{6!}{(6-3)!}=\frac{6!}{3!}=6 \cdot 5 \cdot 4=120\end{align*}.

## Permutations and Graphing Calculators

Most graphing calculators have the ability to calculate a permutation.

Evaluate 6P3\begin{align*}_6P_3\end{align*} using a graphing calculator.

Type the first value of the permutation, the n\begin{align*}n\end{align*}. Choose the [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 k\begin{align*}k\end{align*}. Press [ENTER] to evaluate.

## Permutations and Probability

The letters of the word HOSPITAL are arranged at random. How many different arrangements can be made? What is the probability that the last letter is a vowel?

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.

76543213=15,120\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 chapter, has the formula:

Probability (success)=number of ways to get successtotal number of possible outcomes\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.

P(last letter is a vowel)=15,12040,320=38\begin{align*}P(last \ letter \ is \ a \ vowel)=\frac{15,120}{40,320}=\frac{3}{8}\end{align*}

Multimedia Link: For more help with permutations, visit the http://regentsprep.org/REgents/math/ALGEBRA/APR2/LpermProb.htm - Algebra Lesson Page by Regents Prep.

## Practice Set

1. Define permutation.

In 2 – 19, evaluate each permutation.

1. 7!
2. 10!
3. 1!
4. 5!
5. 9!
6. 3!
7. 4!+4!\begin{align*}4!+4!\end{align*}
8. 16!5!\begin{align*}16!-5!\end{align*}
9. 98!96!\begin{align*}\frac{98!}{96!}\end{align*}
10. 11!2!\begin{align*}\frac{11!}{2!}\end{align*}
11. 301!300!\begin{align*}\frac{301!}{300!}\end{align*}
12. 8!3!\begin{align*}\frac{8!}{3!}\end{align*}
13. 2!+9!\begin{align*}2!+9!\end{align*}
14. 11P2\begin{align*}_{11} P_2\end{align*}
15. 5P5\begin{align*}_5P_5\end{align*}
16. 5P3\begin{align*}_5P_3\end{align*}
17. 15P10\begin{align*}_{15}P_{10}\end{align*}
18. 60P59\begin{align*}_{60}P_{59}\end{align*}
19. How many ways can 14 books be organized on a shelf?
20. How many ways are there to choose 10 objects, four at a time?
21. How many ways are there to choose 21 objects, 13 at a time?
22. A running track has eight lanes. In how many ways can 8 athletes be arranged to start a race?
23. Twelve horses run a race.
1. How many ways can first and second places be won?
2. How many ways will all the horses finish the race?
24. Six actors are waiting to audition. How many ways can the director choose the audition schedule?
25. Jerry, Kerry, Larry, and Mary are waiting at a bus stop. What is the probability that Mary will get on the bus first?
26. How many permutations are there of the letters in the word “HEART”?
27. How many permutations are there of the letters in the word “AMAZING”?
28. 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?
29. What is the probability you choose two cards (without replacement) from a standard 52-card deck and both cards are jacks?
30. 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:
1. The town cannot host a Super Bowl two consecutive years?
2. The town can host a Super Bowl two consecutive years?

Mixed Review

1. Graph the solution to the following system: 2x3y>9y<1\begin{align*}& 2x-3y > -9\\ & y<1\end{align*}
2. Convert 24 meters/minute to feet/second.
3. Solve for t:|t6|14\begin{align*}t: |t-6| \le -14\end{align*}.
4. Find the distance between 6.15 and –9.86.
5. Which of the following vertices provides the minimum cost according to the equation 12x+20y=cost: (3,6),(9,0),(6,2),(0,11)\begin{align*}12x+20y=cost: \ (3,6),(9,0),(6,2),(0,11)\end{align*}?
6. Write the system of inequalities pictured below.

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