1 / 24

240 likes | 539 Vues

Counting. 11.3. Apply the fundamental counting principle Calculate and apply permutations Calculate and apply combinations. Tree Diagram.

Télécharger la présentation
## Counting

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Counting**11.3 Apply the fundamental counting principle Calculate and apply permutations Calculate and apply combinations**Tree Diagram**Suppose that a quiz has only two questions. The first is a multiple-choice question withfour choices, A, B, C, or D, and the second is a true-false (T-F) question. A tree diagram can be used to count the ways that this quiz can be answered.**Tree Diagram**A tree diagram is a systematic way of listing every possibility. Wecan see that there are eight ways to answer the test. They areAT, AF, BT, BF, CT, CF, DT, and DF.For instance, CF indicates a quiz with answers of C on the first question and F on the second question.**Fundamental Counting Principle**Let E1, E2, E3,…, Enbe a sequence of n independent events. If event Ek can occur mk ways for k=1, 2, 3,…, n,then there are m1•m2•m3•… •mn ways for all n events to occur. Two events are independent if neither event influences the outcome of the other.**An exam contains four true-false questions and six**multiple-choice questions. Each multiple-choice question has five possible answers. Count the number of ways that the exam can be answered. Solution This is a sequence of ten independent events. There are two ways to answer each of the first four questions. There are five ways to answer the next six questions. Example: Counting ways to answer an exam**The number of ways to answer the exam is**Example: Counting ways to answer an exam**Sometimes a license plate is limited to 3 uppercase letters**(A through Z) followed by 3 digits (0 through 9). For example, ABB 112 would be a valid license plate. Would this formatprovide enough license plates for a state with 8 million vehicles? Example: Counting license plates**26 choices for letters**10 choices for digits There are 26 ways to choose each of the three letters and 10 ways to choose each of the 3 digits. This format for license plates could accommodate more than 8 million vehicles. Example: Counting license plates**Count the total number of 800 numbers if the local portion**of a telephone number (the last seven digits) does not start with a 0 or 1. Solution A toll-free 800 number assumes the following form. Example: Counting toll-free telephone numbers**We can think of choosing the remaining digits for the local**number as seven independent events. Since the local number cannot begin with a 0 or 1, there are eight possibilities(2 to 9) for the first digit. The remaining six digits can be any number from 0 to 9, sothere are ten possibilities for each of these digits. The total is given by Example: Counting toll-free telephone numbers**n-Factorial**For any natural number n, n! = n(n – 1)(n – 2) … (3)(2)(1) and 0! = 1.**Compute n! for and n = 0, 1, 2, 3, 4,5 by hand.Use a**calculator to find 8!, 13!, and 25!. Solution 0! = 11! = 1 2! = 2 1 = 2 3! = 3 2 1 = 6 4! = 4 3 2 1 = 24 5! = 5 4 3 2 1 = 120 Example: Calculating factorials**Here’s the calculator display. Note that the value for 25!**is an approximation. Notice how rapidly n! increases! Example: Calculating factorials**Permutation**A permutation is an ordering or arrangement.**Permutations of n ElementsTaken r at a Time**If P(n, r)denotes the number of permutations of n elements taken r at a time, withr≤ n,then**For a class of 30 students, how many arrangements are there**in which 4 students each givea speech? Solution The number of permutations of 30 elements taken 4 at a time is given by There are 657,720 ways to arrange the four speeches. Example: Calculating permutations**Combination**A combination is not an ordering or arrangement, but rather a subset of a set of elements. Order is not important when finding combinations. The number of possible subsets, or combinations, is denoted either C(4, 2) or**Combinations of n ElementsTaken r at a Time**If C(n, r)denotes the number of combinations of n elements taken r at a time, withr≤ n,then**Calculate each of the following. Support your answer by**using a calculator. Solution Example: Calculating C(n,r)**A college student has five courses left in her major and**plans to take two of them this semester.Assuming that this student has the prerequisites for all five courses, determine how manyways these two courses can be selected. Example: Counting combinations**Solution**The order in which the courses are selected is unimportant. From a set of 5courses, the student selects a subset of 2 courses. The number of subsets is There are 10 ways to select two courses from a set of five. Example: Counting combinations**How many committees of six people can be selected from six**women and three men, if a committee must consist of at least two men? Solution Two Men: Committee would be two men and four women. Example: Counting committees**Three Men: Committee would be three men and three women**The total number of possible committees would be45 + 20 = 65. Example: Counting committees

More Related