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9-6 Permutations and Combinations 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: "9-6 Permutations and Combinations Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation."— Presentation transcript:

1 9-6 Permutations and Combinations 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 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 3. How many different 4–digit phone extensions are possible? 16 10, Course Permutations and Combinations

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

4 Learn to find permutations and combinations. Course Permutations and Combinations

5 Vocabulary factorial permutation combination Insert Lesson Title Here Course Permutations and Combinations

6 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

7 Evaluate each expression. Additional Example 1A & 1B: Evaluating Expressions Containing Factorials Course Permutations and Combinations A. 8! = 40,320 8! 6! Write out each factorial and simplify. 8 7 = 56 B. Multiply remaining factors.

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

9 Evaluate each expression. Try This: Example 1A & 1B Course Permutations and Combinations A. 10! = 3,628,800 7! 5! Write out each factorial and simplify. 7 6 = 42 B. Multiply remaining factors.

10 Try This: Example 1C Course Permutations and Combinations Subtract within parentheses = 504 9! 6!  5  4  3  2  1 C. 9! (8 – 2)!

11 Course Permutations and Combinations 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

12 Course Permutations and Combinations If no letter can be used more than once, there are 60 permutations 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 Notice that the product can be written as a quotient of factorials.  60 = = = 60 permutations = 5! 2!

13 Course Permutations and Combinations

14 Jim has 6 different books. Additional Example 2A: Finding Permutations Course Permutations and Combinations A. Find the number of orders in which the 6 books can be arranged on a shelf ! (6 – 6)! = 6! 0! = = 6 P 6 = The number of books is 6. The books are arranged 6 at a time. There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

15 Additional Example 2B: Finding Permutations Course Permutations and Combinations B. 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 ! (6 – 3)! = 6! 3! = = 6 P 3 = The number of books is 6. The books are arranged 3 at a time. = 120

16 Course Permutations and Combinations = ! (7 – 7)! = 7! 0! = P 7 = The number of cans is 7. The cans are arranged 7 at a time. There are 5040 orders in which to arrange 7 soup cans. Try This: Example 2A A. 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.

17 Course Permutations and Combinations There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways. = ! (7 – 4)! = 7! 3! = P 4 = The number of cans is 7. The cans are arranged 4 at a time. = 840 There are 7 soup cans in the pantry. Try This: Example 2B B. If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged.

18 Course Permutations and Combinations A combination is a selection of things in any order.

19 Course Permutations and Combinations If no letter is used more than once, there is only 1 combination of the first 3 letters of the alphabet. ABC, ACB, BAC, BCA, CAB, and CBA are considered to be the same combination of A, B, and C because the order does not matter. If no letter is used more than once, there are 10 combinations of the first 5 letters of the alphabet, when taken 3 at a time. To see this, look at the list of permutations on the next slide.

20 Course Permutations and Combinations 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 =

21 Course Permutations and Combinations

22 Additional Example 3A: Finding Combinations Course Permutations and Combinations Mary wants to join a book club that offers a choice of 10 new books each month. A. If Mary wants to buy 2 books, find the number of different pairs she can buy. 10 possible books 2 books chosen at a time 10! 2!(10 – 2)! = 10! 2!8! = (2 1)( ) 10 C 2 = = 45 There are 45 combinations. This means that Mary can buy 45 different pairs of books.

23 Additional Example 3B: Finding Combinations Course Permutations and Combinations B. If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy. 10 possible books 7 books chosen at a time 10! 7!(10 – 7)! = 10! 7!3! 10 C 7 = ( )(3 2 1) = = 120 There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

24 Try This: Example 3A Course Permutations and Combinations Harry wants to join a DVD club that offers a choice of 12 new DVDs each month. A. If Harry wants to buy 4 DVDs, find the number of different sets he can buy. 12 possible DVDs 4 DVDs chosen at a time 12! 4!(12 – 4)! = 12! 4!8! = ( )( ) 12 C 4 = = 495

25 Try This: Example 3A Continued Course Permutations and Combinations There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

26 Try This: Example 3B Course Permutations and Combinations B. If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy. 12 possible DVDs 11 DVDs chosen at a time 12! 11!(12 – 11)! = 12! 11!1! = ( )(1) 12 C 11 = = 12

27 Try This: Example 3B Continued Course Permutations and Combinations There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

28 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? 4. A group of 12 people are forming a committee. How many different 4-person committees can be formed? Lesson Quiz ,880 Insert Lesson Title Here 40, Course Permutations and Combinations 9! 5!


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