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15.2 Counting Methods: Permutations

15.2 Counting Methods: Permutations. ©2002 by R. Villar All Rights Reserved. Permutations. meat bread sandwich. white wheat rye. ham on white ham on wheat ham on rye. In order to find probabilities, you must be able to count the total number of outcomes.

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15.2 Counting Methods: Permutations

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  1. 15.2 Counting Methods: Permutations ©2002 by R. Villar All Rights Reserved

  2. Permutations meat bread sandwich white wheat rye ham on white ham on wheat ham on rye In order to find probabilities, you must be able to count the total number of outcomes. For example, a certain deli offers 3 types of meat (ham, turkey, and roast beef) and 3 types of bread (white, wheat, and rye). How many choices do the customers have for a meat sandwich? One way to determine this is by using a tree diagram... ham turkey roast beef white wheat rye turkey on white turkey on wheat turkey on rye white wheat rye roast beef on white roast beef on wheat roast beef on rye There are 9 choices (or possible outcomes).

  3. Another way to count the number of possible sandwiches is to use the Fundamental Counting Principle. Because you have 3 choices for meat and 3 choices for bread… …the total number of choices is 3 • 3 = 9. Fundamental Counting Principle If one event can occur m ways and another can occur in n ways, then the number of ways that both events can occur is m • n. This principle can be extended to three or more events. For example, if three events can occur m, n, and p ways, then the number of ways that all three can occur is m • n • p

  4. Example: The standard configuration for a New York license plate is 3 digits followed by 3 letters. How many different license plates are possible if digits and letters can be repeated? There are 10 choices for each digit and 26 choices for each letter. You can use the Fundamental Counting Principle to find the number of license plates... Number of Plates = 10 • 10 • 10 • 26 • 26 • 26 =17,576,000 different license plates How many license plates are possible if digits and letters cannot be repeated? Since digits cannot be repeated, there are still 10 choices for the first digit, but only 9 for the second digit, and 8 for the third digit. The letters work similarly... Number of Plates = 10 • 9 • 8 • 26 • 25 • 24 =11,232,000 different license plates

  5. Each of the licenses in the previous examples is called a permutation. For example, there are six permutations of the letters A, B, and C: ABC, ACB, BAC, BCA, CAB, CBA The Fundamental Counting Principle can be used to determine the number of permutations of n objects. For example, you can find the number of ways you can arrange the letters A, B, and C by multiplying. There are 3 choices for the first letter, 2 choices for the second letter, and 1 choice for the third letter, so there are 3 • 2 • 1 = 6 ways to arrange the letters. In general, the number of permutations ofn distinct objects is: n! = n • (n – 1) • (n – 2) • … 4 • 3 • 2 • 1

  6. Example: Twelve skiers are competing in the final round of the Olympic freestyle skiing aerial competition. A. In how many different ways can the skiers finish the competition? (Assume there are no ties) There are 12! different ways that the skiers can finish. 12! = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 =479,001,600 different ways B. In how many different ways can the skiers finish the competition? (Assume there are no ties) Any of the 12 skiers can finish first, then any of the 11 can finish second, and any of the remaining 10 can finish third… 12 • 11 • 10 =1320 different ways

  7. Probabilities Involving Permutations or Combinations 1 1 P(playing 8 in order) = = ≈ 0.0000248 8! 40, 320 You put a CD that has 8 songs in your CD player. You set the player to play the songs at random. The player plays all 8 songs without repeating any song. What is the probability that the songs are played in the same order they are listed on the CD? SOLUTION There are 8! differentpermutations of the 8 songs. Of these, only 1 is the order in which the songs are listed on the CD. So, the probability is:

  8. Probabilities Involving Permutations or Combinations 6 3 P(playing 2 favorites first) = = = ≈ 0.214 28 14 You put a CD that has 8 songs in your CD player. You set the player to play the songs at random. The player plays all 8 songs without repeating any song. You have 4 favorite songs on the CD. What is the probability that 2 of your favorite songs are played first, in any order? SOLUTION There are different combinations of 2 songs. Of these, contain 2 of your favorite songs. So, the probability is:

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