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10-9 Permutations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation.

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Presentation on theme: "10-9 Permutations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation."— Presentation transcript:

1 10-9 Permutations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

2 Warm Up Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna 16 Course Permutations

3 Warm Up Find the number of possible outcomes. 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 12 Course Permutations

4 Warm Up Find the number of possible outcomes. 3. How many different 4–digit phone extensions are possible? 10,000 Course Permutations

5 Problem of the Day What is the probability that a 2-digit whole number will contain exactly one 1? Course Permutations 17 90

6 Learn to find permutations. Course Permutations

7 Vocabulary factorial permutation Insert Lesson Title Here Course Permutations

8 Course Permutations and Combinations The factorial of a number is the product of all the whole numbers from the number down to 1. The factorial of 0 is defined to be 1. 5! ! = Read 5! as five factorial. Reading Math

9 Evaluate each expression. Example 1: Evaluating Expressions Containing Factorials Course Permutations A. 9! = 362,880 8! 6! Write out each factorial and simplify. 8 7 = 56 B. Multiply remaining factors.

10 Example 1: Evaluating Expressions Containing Factorials Course Permutations and Combinations Subtract within parentheses = ! 7! C. 10! (9 – 2)!

11 Evaluate each expression. Check It Out: Example 1 Course Permutations A. 10! = 3,628,800 7! 5! Write out each factorial and simplify. 7 6 = 42 B. Multiply remaining factors.

12 Check It Out: Example 1 Course Permutations Subtract within parentheses = 504 9! 6! C. 9! (8 – 2)!

13 Course Permutations A permutation is an arrangement of things in a certain order. If no letter can be used more than once, there are 6 permutations of the first 3 letters of the alphabet: ABC, ACB, BAC, BCA, CAB, and CBA. first letter ? second letter ? third letter ? 3 choices2 choices1 choice The product can be written as a factorial = 3! = 6

14 Course Permutations If no letter can be used more than once, there are 60 permutations (orders) of the first 5 letters of the alphabet, when taken 3 at a time: ABE, ACD, ACE, ADB, ADC, ADE, and so on. first letter ? second letter ? third letter ? 5 choices4 choices3 choices = 60 permutations

15 Course Permutations ABCABDABEACDACEADEBCDBCEBDECDE ACBADBAEBADCAECAEDBDCBECBEDCED BACBADBAECADCAEDAECBDCBEDBEDCE BCABDABEACDACEADEADBCCEBDEBDEC CABDABEABDACEACEADDCBEBCEBDECD CBADBAEBADCAECAEDADBCECBEDBEDC These 6 permutations are all the same combination. In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is =

16 Jim has 6 different books. Example 2A: Finding Permutations Course Permutations Find the number of orders in which the 6 books can be arranged on a shelf = There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

17 Course Permutations = ! = Use 7! There are 5040 orders in which to arrange 7 soup cans. Check It Out: Example 2A Find the number of orders in which all 7 soup cans can be arranged on a shelf. There are 7 soup cans in the pantry.

18 Example 2B: Finding Permutations Course Permutations If the shelf has room for only 3 of the books, find the number of ways 3 of the 6 books can be arranged. There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways = 6P36P3 The number of books is 6. The books are arranged 3 at a time. = 120

19 Course Permutations There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways. = P47P4 The number of cans is 7. The cans are arranged 4 at a time. = 840 There are 7 soup cans in the pantry. Check It Out: Example 2B If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.

20 Evaluate each expression. 1. 9! There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race? Lesson Quiz ,880 Insert Lesson Title Here 40,320 Course Permutations 9! 5!


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