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

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

10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3 10-9 Permutations Warm Up Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna 16

10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3 10-9 Permutations Warm Up Find the number of possible outcomes. 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 12

10-9 Permutations Warm Up Find the number of possible outcomes.
Course 3 10-9 Permutations Warm Up Find the number of possible outcomes. 3. How many different 4–digit phone extensions are possible? 10,000

10-9 Permutations Problem of the Day
Course 3 10-9 Permutations Problem of the Day What is the probability that a 2-digit whole number will contain exactly one 1? 17 90

Course 3 10-9 Permutations Learn to find permutations.

Insert Lesson Title Here
Course 3 10-9 Permutations Insert Lesson Title Here Vocabulary factorial permutation

5! = 5 • 4 • 3 • 2 • 1 10-9 Permutations and Combinations
Course 3 10-9 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! = 5 • 4 • 3 • 2 • 1 Read 5! as “five factorial.” Reading Math

Example 1: Evaluating Expressions Containing Factorials
Course 3 10-9 Permutations Example 1: Evaluating Expressions Containing Factorials Evaluate each expression. A. 9! 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362,880 8! B. 6! Write out each factorial and simplify. 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1 Multiply remaining factors. 8 • 7 = 56

Example 1: Evaluating Expressions Containing Factorials
Course 3 10-9 Permutations and Combinations Example 1: Evaluating Expressions Containing Factorials 10! (9 – 2)! C. 10! 7! Subtract within parentheses. 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7  6  5  4  3  2  1 10 • 9 • 8 = 720

10-9 Permutations Check It Out: Example 1 Evaluate each expression.
Course 3 10-9 Permutations Check It Out: Example 1 Evaluate each expression. A. 10! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 3,628,800 7! B. 5! Write out each factorial and simplify. 7 • 6 • 5 • 4 • 3 • 2 • 1 5 • 4 • 3 • 2 • 1 Multiply remaining factors. 7 • 6 = 42

10-9 Permutations Check It Out: Example 1 9! C. (8 – 2)! 9!
Course 3 10-9 Permutations Check It Out: Example 1 9! (8 – 2)! C. 9! 6! Subtract within parentheses. 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6  5  4  3  2  1 9 • 8 • 7 = 504

Course 3 10-9 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 choices 2 choices 1 choice The product can be written as a factorial. 3 • 2 • 1 = 3! = 6

Course 3 10-9 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 choices 4 choices 3 choices = 60 permutations 5 • 4 • 3

10-9 Permutations ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE
Course 3 10-9 Permutations ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ACB ADB AEB ADC AEC AED BDC BEC BED CED BAC BAD BAE CAD CAE DAE CBD CBE DBE DCE BCA BDA BEA CDA CEA DEA DBC CEB DEB DEC CAB DAB EAB DAC EAC EAD DCB EBC EBD ECD CBA DBA EBA DCA ECA EDA DBC ECB EDB EDC 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 = 10. 60 6

Example 2A: Finding Permutations
Course 3 10-9 Permutations Example 2A: Finding Permutations Jim has 6 different books. Find the number of orders in which the 6 books can be arranged on a shelf. 6 • 5 • 4 • 3 • 2 • 1 = 720 There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

10-9 Permutations Check It Out: Example 2A
Course 3 10-9 Permutations Check It Out: Example 2A There are 7 soup cans in the pantry. Find the number of orders in which all 7 soup cans can be arranged on a shelf. Use 7! 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1 = 5040 There are 5040 orders in which to arrange 7 soup cans.

Example 2B: Finding Permutations
Course 3 10-9 Permutations Example 2B: Finding 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. The number of books is 6. = 6P3 6 • 5 • 4 The books are arranged 3 at a time. = 120 There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

10-9 Permutations Check It Out: Example 2B
Course 3 10-9 Permutations Check It Out: Example 2B There are 7 soup cans in the pantry. If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged. The number of cans is 7. = 7 • 6 • 5 • 4 7P4 The cans are arranged 4 at a time. = 840 There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

Insert Lesson Title Here
Course 3 10-9 Permutations Insert Lesson Title Here Lesson Quiz Evaluate each expression. 1. 9! 2. 3. There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race? 362,880 9! 5! 3024 40,320

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