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Warm Up Evaluate. 1. 5  4  3  2  1 2. 7  6  5  4  3  2  1 3.4. 5.6. 120 5040 4210 10 70.

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Presentation on theme: "Warm Up Evaluate. 1. 5  4  3  2  1 2. 7  6  5  4  3  2  1 3.4. 5.6. 120 5040 4210 10 70."— Presentation transcript:

1 Warm Up Evaluate. 1. 5  4  3  2  1 2. 7  6  5  4  3  2  1 3.4. 5.6. 120 5040 4210 10 70

2 You have previously used tree diagrams to find the number of possible combinations of a group of objects. In this lesson, you will learn to use the Fundamental Counting Principle.

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4 Example 1A: Using the Fundamental Counting Principle To make a yogurt parfait, you choose one flavor of yogurt, one fruit topping, and one nut topping. How many parfait choices are there? Yogurt Parfait (choose 1 of each) Flavor Plain Vanilla Fruit Peaches Strawberries Bananas Raspberries Blueberries Nuts Almonds Peanuts Walnuts

5 Example 1A Continued number of flavors times number of fruits number of nuts times equals number of choices 2  5  3 = 30 There are 30 parfait choices.

6 Example 1B: Using the Fundamental Counting Principle A password for a site consists of 4 digits followed by 2 letters. The letters A and Z are not used, and each digit or letter many be used more than once. How many unique passwords are possible? digit digit digit digit letter letter 10  10  10  10  24  24 = 5,760,000 There are 5,760,000 possible passwords.

7 Check It Out! Example 1a A “make-your-own-adventure” story lets you choose 6 starting points, gives 4 plot choices, and then has 5 possible endings. How many adventures are there? number of starting points  number of plot choices number of possible endings  = number of adventures 6  4  5 = 120 There are 120 adventures.

8 Check It Out! Example 1b A password is 4 letters followed by 1 digit. Uppercase letters (A) and lowercase letters (a) may be used and are considered different. How many passwords are possible? Since both upper and lower case letters can be used, there are 52 possible letter choices. letter letter letter letter number 52  52  52  52  10 = 73,116,160 There are 73,116,160 possible passwords.

9 A permutation is a selection of a group of objects in which order is important. There is one way to arrange one item A. A second item B can be placed first or second. A third item C can be first, second, or third for each order above. 1 permutation 2 · 1 permutations 3 · 2 · 1 permutations EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

10 You can see that the number of permutations of 3 items is 3 · 2 · 1. You can extend this to permutations of n items, which is n · (n – 1) · (n – 2) · (n – 3) ·... · 1. This expression is called n factorial, and is written as n!. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

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12 Sometimes you may not want to order an entire set of items. Suppose that you want to select and order 3 people from a group of 7. One way to find possible permutations is to use the Fundamental Counting Principle. First Person Second Person Third Person There are 7 people. You are choosing 3 of them in order. 7 choices 6 choices 5 choices  = 210 permutations EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

13 arrangements of 4 4! 4 · 3 · 2 · 1 Another way to find the possible permutations is to use factorials. You can divide the total number of arrangements by the number of arrangements that are not used. In the previous slide, there are 7 total people and 4 whose arrangements do not matter. arrangements of 7 = 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 210 This can be generalized as a formula, which is useful for large numbers of items. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

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15 Example 2A: Finding Permutations How many ways can a student government select a president, vice president, secretary, and treasurer from a group of 6 people? This is the equivalent of selecting and arranging 4 items from 6. = 6 5 4 3 = 360 Divide out common factors. There are 360 ways to select the 4 people. Substitute 6 for n and 4 for r in EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

16 Example 2B: Finding Permutations How many ways can a stylist arrange 5 of 8 vases from left to right in a store display? Divide out common factors. = 8 7 6 5 4 = 6720 There are 6720 ways that the vases can be arranged. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

17 Check It Out! Example 2a Awards are given out at a costume party. How many ways can “most creative,” “silliest,” and “best” costume be awarded to 8 contestants if no one gets more than one award? = 8 7 6 = 336 There are 336 ways to arrange the awards. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

18 Check It Out! Example 2b How many ways can a 2-digit number be formed by using only the digits 5–9 and by each digit being used only once? = 5 4 = 20 There are 20 ways for the numbers to be formed. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

19 A combination is a grouping of items in which order does not matter. There are generally fewer ways to select items when order does not matter. For example, there are 6 ways to order 3 items, but they are all the same combination: 6 permutations  {ABC, ACB, BAC, BCA, CAB, CBA} 1 combination  {ABC} EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

20 To find the number of combinations, the formula for permutations can be modified. Because order does not matter, divide the number of permutations by the number of ways to arrange the selected items. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

21 Permutations A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated? Warm Up: EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

22 Three Types of Permutations 1. Distinct with repetition. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations? Let's look at a 3 combination lock with numbers 0 through 9 There are 10 choices for the first number There are 10 choices for the second number and you can repeat the first number There are 10 choices for the third number and you can repeat

23 Three Types of Permutations 2. Distinct without repetition. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations? Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition). 1 st 2 nd 3 rd 4 th

24 EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

25 When deciding whether to use permutations or combinations, first decide whether order is important. Use a permutation if order matters and a combination if order does not matter. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

26 You can find permutations and combinations by using nPr and nCr, respectively, on scientific and graphing calculators. Helpful Hint EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

27 Example 3: Application There are 12 different-colored cubes in a bag. How many ways can Randall draw a set of 4 cubes from the bag? Step 1 Determine whether the problem represents a permutation of combination. The order does not matter. The cubes may be drawn in any order. It is a combination. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

28 Example 3 Continued = 495 Divide out common factors. There are 495 ways to draw 4 cubes from 12. 5 Step 2 Use the formula for combinations. n = 12 and r = 4 EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

29 Check It Out! Example 3 The swim team has 8 swimmers. Two swimmers will be selected to swim in the first heat. How many ways can the swimmers be selected? = 28 The swimmers can be selected in 28 ways. 4 Divide out common factors. n = 8 and r = 2 EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

30 Combinations To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

31 Combinations A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

32 Three Types of Permutations 3. Involving n objects that are not distinct. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations? Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition). RRRAAEENG

33 The third type of permutation is involving n objects that are not distinct. How many different combinations of letters in specific order (but not necessarily English words) can be formed using ALL the letters in the word REARRANGE? The "words" we form will have 9 letters so we need 9 spots to place the letters. Notice some of the letters repeat. We need to use R 3 times, A 2 times, E 2 times and N and G once. First we choose positions for the R's. There are 9 positions and we'll choose 3, order doesn't matter 9C39C3 That leaves 6 positions for 2 A's. RRR  6 C 2 AA That leaves 4 positions for 2 E's.  4C2 4C2 That leaves 2 positions for the N. EE  2C1 2C1 That leaves 1 position for the G. NG  1 C 1 representative example 84  15  6  2  1 = 15 120 possible "words" Not Examinable.. Just for Fun

34 Combinations A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Center: Forwards:Guards: Thus, the number of ways to select the starting line up is 2*10*6 = 120. EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?

35 Lesson Quiz 1. Six different books will be displayed in the library window. How many different arrangements are there? 2. The code for a lock consists of 5 digits. The last number cannot be 0 or 1. How many different codes are possible? 80,000 720 3. The three best essays in a contest will receive gold, silver, and bronze stars. There are 10 essays. In how many ways can the prizes be awarded? 4. In a talent show, the top 3 performers of 15 will advance to the next round. In how many ways can this be done? 455 720 EQ: How do we solve problems involving the Fundamental Counting Principle, and permutations and combinations?


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