1. Counting arrangements - Permutations
Chapter 2-2
Counting arrangements - Permutations

2. Counting Arrangements
The idea of counting arrangements is to find the number of different ways to arrange a set of things.
For example take the set {1,2,3}. There are 6 ways to arrange this set into a 3 digit number without using a number twice.
123, 132, 213, 231, 312, 321
Each of these arrangements is called a “permutation”

3. Permutation
Order of the arrangement matters! That means 123 is a different arrangement than 321.
Also, in a permutation, you cannot use an element twice (Like 331)
A permutation is an ordered list of elements selected from a set. Two permutations are different unless they consist of exactly the same elements in exactly the same order.

4. Example
A program committee consist of 5 members: Audrey(A), Bill(B), Connie(C), David(D), and Ellie(E). The committee is planning the next event, and one of its members is to handle publicity and another is to reserve a room for the program. Find the number of different ways of assigning pairs of committee members to the two tasks of publicity and room arrangements.
Solution:
There are 5 ways to assign the first task. Once that person is selected, there are 4 ways to assign the next task. 5 X 4 = 20

5. Example 2
How many permutations of 3 letters can be formed with letters selected from a set of 6 distinct letters?
Solution:
There are 6 ways to assign the first letter, 5 ways to assign the second, 4 ways to assign the third.
6 X 5 X 4 = 120

6. Permutation Formula
Permutation Principle: The number of permutations of r objects selected from a set of n distinct objects is 𝑛× 𝑛−1 …×(𝑛−𝑟+1)
n: number of things you are choosing from
r: number you are choosing
𝑃 𝑛,𝑟 = 𝑛! 𝑛−𝑟 ! “!” called “factorial” From the previous example: 6,3 = 6! 6−3 !

7. Example 3
A map showing the 4 states of Arizona, Colorado, New Mexico, and Utah, is to be colored so each state is a different color. There are 8 colors available. In how many different ways could the map be colored?
𝑃 8,4 = 8! 8−4 ! =8×7×6×5=1680

8. Example 4
Lets consider the word “infinite.” How many 8-letter arrangements can be formed from the letters of this word?
Solution:
𝑃 8,8 = 8! 8−8 ! =8×7×6×5×4×3×2×1=40320
!2! =3360
3 “i’s”
2 “n’s”

9. 1
Evaluate 𝑃 7,2

10. 2
Evaluate 𝑃 9,5

11. 3
A traveler plans to visit 4 of the cities Athens, Berlin, Cairo, Dusseldorf, Edinburgh, and Florence. An itinerary is a list of 4 of the cities to be visited, in order. How many different itineraries are there?

12. 4
A conductor has 5 songs to conduct during a concert, and he can conduct them in any order. In how many ways can he organize the concert?

13. 5
How many 3-digit numbers can be formed with the digits 2, 4, 6, 8, and 9 if each digit is used at most once? How many of these numbers are smaller than 500?

14. 6
How many 2 letter “words” can be formed from the letters of “consider”?

15. 7
A product code is formed by arranging 5 different letters from the collection 𝐵,𝐻,𝑄,𝑇,1,𝑍 one after another. How many different such product codes are there? How many of them containing the letter H?

16. 1
An experiment with 5 outcomes has 𝑤 1 =.23, 𝑤 2 =.14, 𝑤 3 = 𝑤 4 , 𝑤 5 =.29. Find 𝑤 4 .

17. 2
An experiment with outcomes 𝑂 1 , 𝑂 2 , 𝑂 3 , 𝑂 4 ,and 𝑂 5 has an assignment of probabilities 𝑤 1 , 𝑤 2 , 𝑤 3 , 𝑤 4 ,and 𝑤 5 . Suppose 𝑂 2 is three times as likely as 𝑂 4 , 𝑤 1 =.4 and 𝑤 3 = 𝑤 4 = 𝑤 5 . Find 𝑤 2 .

18. 3
Let 𝑆= 𝑂 1 , 𝑂 2 , 𝑂 3 , 𝑂 4 be a sample space. Suppose Pr 𝑂 1 , 𝑂 3 =.55 and Pr 𝑂 1 , 𝑂 2 , 𝑂 4 =.75. Find Pr 𝑂 2 , 𝑂

19. 4
Suppose an experiment has 4 outcomes and the frequencies of the outcomes are 3 𝑛 , 11 𝑛 , 5 𝑛 , 7 𝑛 . What does n have to be for these frequencies to reflect probabilities?

20. 5
An experiment has outcomes 𝑂 1 , 𝑂 2 ,and 𝑂 3 , with Pr 𝑂 1 , 𝑂 2 =0.6, Pr 𝑂 1 , 𝑂 3 =0.8. Find Pr⁡[ 𝑂 2 ].

21. 6
An experiment has outcomes 𝑂 1 , 𝑂 2 ,and 𝑂 3 , with Pr 𝑂 1 , 𝑂 2 =0.75, Pr 𝑂 1 , 𝑂 3 = Find Pr⁡[ 𝑂 1 ].

22. 7
Consider a die constructed so that outcomes of 1, 2, 3, or 4 dots are equally likely, 5 is twice as likely as 1, and 6 is twice as likely as 5. What probabilities should be assigned to each outcome?

23. 8
We have a 6-sided die (3 different odd number sides and 3 different even number sides). It is weighted in such fashion that: 1. the probabilities for all the odd sides are the same. 2. The probabilities for all the even sides are the same. The probability of obtaining a given odd side is twice that of obtaining a given even side. What is the probability of obtaining a given odd side?

24. 9
We have a 6-sided die (3 different odd number sides and 3 different even number sides). It is weighted in such fashion that: 1. the probabilities for all the odd sides are the same. 2. The probabilities for all the even sides are the same. The probability of obtaining a given odd side is twice that of obtaining a given even side. What is the probability of obtaining a 1, 3, or 6?

25. 10
We have a 10-sided die (5 different odd number sides and 5 different even number sides). It is weighted in such fashion that: 1. the probabilities for all the odd sides are the same. 2. The probabilities for all the even sides are the same. The probability of obtaining a given odd side is twice that of obtaining a given even side. What is the probability of obtaining a given odd side?