Section 3-4 Permutations

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Presentation transcript:

Section 3-4 Permutations A permutation is an ordered arrangement of objects. The number of different permutations of n distinct object is n! The ! Symbol means “factorial” and indicates that you start with the number given and multiply that times every number between that number and zero. 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880 The calculator will do factorials for you; Math – Prb – 4 accesses that feature. To find the number of ways that a permutation can occur, use the TI-84. Math – PRB – 2 Enter the total number of items in your list. Press Math - PRB – 2 and Enter Enter the number of items you wish to order and press enter.

Section 3-4 Distinguishable Permutations A distinguishable permutation must be done by hand. 𝑛! 𝑛 1 ! 𝑛 2 ! 𝑛 3 !… 𝑛 𝑘 ! Alpha and y= on your calculator will allow you to enter this as a fraction, making it easier. Again, the calculator will also let you enter the factorial symbol, saving time and reducing the chance of making a simple error in entering the numbers.

Section 3-4 Combinations A combination is a selection of r objects from a group of n objects without regard to order. To find the number of ways that a combination can occur, use the TI-84. Math – PRB – 3 Enter the total number of items in your list. Press Math - PRB – 3 and Enter Enter the number of items you wish to select and press enter.

Section 3-4 Example 1 The objective of a 9 x 9 Sudoku number puzzle is to fill the grid so that each row, each column, and each 3 x 3 grid contains the digits 1 to 9, with no repeats. How many different ways can the first row of a blank 9 x 9 Sudoku grid be filled? SOLUTION: There are 9 digits that could fill the first spot, 8 for the second spot, and so forth. This can be expressed as either 9!, or as 9P9. 9!=362,880 9P9 also equals 362,880. So, there are 362,880 ways to fill the first row of a 9 x 9 Sudoku number puzzle.

Section 3-4 Example 2 Find the number of ways of forming three-digit codes in which no digit is repeated. SOLUTION: To form a three-digit code with no repeating digits, you need to select 3 digits from a group of 10, so n = 10 and r = 3. 10P3 = 720. You could also simply say that you had 10 choices for the first digit, 9 choices for the second digit, and 8 choices for the third digit. 10 * 9 * 8 = 720.

Section 3-4 Example 3 Forty-three race cars started the 2007 Daytona 500. How many ways can the cars finish first, second, and third? SOLUTION: You want to order 3 cars out of 43. This is a permutation. 43P3 = 74,046 43 * 42* 41 = 74,046 43 cars could win the race. Once the winner has won, there are 42 left to fight for 2nd. Once 2nd has been determined, there are 41 left to battle it out for 3rd.

Section 3-4 Example 4 A building contractor is planning to develop a sub-division. The sub-division is to consist of 6 one-story houses, 4 two-story houses and 2 split-level houses. In how many distinguishable ways can the houses be arranged? SOLUTION: This is a distinguishable permutation, the formula for which is 𝑛! 𝑛 1 ! 𝑛 2 ! 𝑛 3 ! 12! 6!4!2! =13,860 different distinguishable ways to arrange the houses.

Section 3-4 Example 5 A state’s department of transportation plans to develop a new section of interstate highway and receives 16 bids for the project. The state plans to hire four of the bidding companies. How many different combinations of four companies can be selected from the 16 bidding companies? SOLUTION: 16C4 = 1820. There are 1820 different combinations of 4 companies that can be selected from the 16 bidding companies.

Section 3-4 Example 6 A student advisory board consists of 17 members. Three members serve as the board’s chair, secretary, and webmaster. Each member is equally likely to serve any of the positions. What is the probability of selecting at random the three members that hold each position? SOLUTION: There are 17P3 different ways that the three positions can be filled. There are 4080 different ways that the three positions can be filled. The chance that you randomly pick the one correct outcome is 1 4080 ≈2.451𝐸 − 4

Section 3-4 Example 7 You have 11 letters consisting of one M, four I’s, four S’s and two P’s. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word Mississippi? SOLUTION: This is a distinguishable permutation question. There are 11! 1!4!4!2! different distinguishable ways to arrange those 11 letters. There are 34,650 different distinguishable ways to arrange those 11 letters. The probability of randomly picking the one arrangement that spells Mississippi is 1 34650 ≈2.886𝐸 − 5

Section 3-4 Example 8 Find the probability of being dealt five diamonds from a standard deck of playing cards. SOLUTION: You need to determine how many ways you can get 5 diamonds, and then how many different 5 card hands are possible. Once you have these values, divide the number of ways to get 5 diamonds by the number of possible hands to find out how likely it is that you get 5 diamonds. 13C5 = 1287 and 52C5 = 2,598,960 P(5 diamonds) = 1287 2,598,960 ≈4.952𝐸 − 4 The probability of getting a diamond flush is approximately 0.0005.

Section 3-4 Example 9 A food manufacturer is analyzing a sample of 400 corn kernels for the presence of a toxin. In this sample, three kernels have dangerously high levels of the toxin. If four kernels are randomly selected from the sample, what is the probability at exactly one kernel contains a dangerously high level of the toxin? SOLUTION: Find the number of possible ways to choose one toxic kernel and three non-toxic kernels, multiply those together and divide the answer by the number of ways to choose 4 kernels. This tells you how likely it is that you will randomly choose exactly one toxic kernel out of four.

Section 3-4 Example 9 SOLUTION: 3C1 = 3 and 397C3 = 10,349,790 3 * 10,349,790 = 31,049,370 There are 31,049,370 different ways to select one toxic kernel out of 400. 400C4 = 1,050,739,900 There are 1,050,739,900 different ways to select 4 kernels out of 400. P(exactly 1 toxic kernel) = 31,049,370 1,050,739,900 ≈0.02955 The probability of getting exactly one toxic kernel is approximately 0.030.

Assignments: Classwork: Page 178 #1–28 Homework: Pages 180–181 #38–52 Evens