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9-6. Permutations and Combinations. Warm Up. Problem of the Day. Lesson Presentation. Pre-Algebra. 9-6. Permutations and Combinations. Pre-Algebra. Warm Up Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna

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9-6

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  1. 9-6 Permutations and Combinations Warm Up Problem of the Day Lesson Presentation Pre-Algebra

  2. 9-6 Permutations and Combinations Pre-Algebra 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 12 10,000

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

  4. Learn to find permutations and combinations.

  5. Vocabulary factorial permutation combination

  6. Reading Math Read 5! as “five factorial.” 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

  7. 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1 Additional Example 1A & 1B: Evaluating Expressions Containing Factorials Evaluate each expression. A. 8! 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 = 40,320 8! B. 6! Write out each factorial and simplify. Multiply remaining factors. 8 •7 = 56

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

  9. 7 • 6 • 5 • 4 • 3 • 2 • 1 5 • 4 • 3 • 2 • 1 Try This: Example 1A & 1B 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. Multiply remaining factors. 7 •6 = 42

  10. 9! (8 – 2)! 9! 6! 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 6  5  4  3  2  1 Try This: Example 1C C. Subtract within parentheses. 9 •8 • 7 = 504

  11. first letter second letter third letter ? ? ? 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. 3 choices 2 choices 1 choice • • The product can be written as a factorial. 3 • 2 • 1 = 3! = 6

  12. first letter second letter third letter ? ? ? 5 • 4 • 3 • 2 • 1 5! 2! = 2 • 1 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. 5 choices 4 choices 3 choices   = 60 permutations Notice that the product can be written as a quotient of factorials. 60 = 5 • 4 • 3 =

  13. 6! (6 – 6)! The number of books is 6. = = 6! The books are arranged 6 at a time. 0! 6 • 5 • 4 • 3 • 2 • 1 = 1 Additional Example 2A: Finding Permutations Jim has 6 different books. A. Find the number of orders in which the 6 books can be arranged on a shelf. 6P6 = 720 There are 720 permutations. This means there are 720 orders in which the 6 books can be arranged on the shelf.

  14. 6! (6 – 3)! The number of books is 6. = = 6! The books are arranged 3 at a time. 3! 6 • 5 • 4 • 3 • 2 • 1 = 3 • 2 • 1 Additional Example 2B: Finding Permutations 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. 6 • 5 • 4 6P3 = = 120 There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

  15. 7! (7 – 7)! The number of cans is 7. = = 7! The cans are arranged 7 at a time. 0! 7 • 6 • 5 • 4 • 3 • 2 • 1 1 Try This: Example 2A There are 7 soup cans in the pantry. A. Find the number of orders in which all 7 soup cans can be arranged on a shelf. 7P7 = = 5040 There are 5040 orders in which to arrange 7 soup cans.

  16. 7! (7 – 4)! The number of cans is 7. = = 7! The cans are arranged 4 at a time. 3! 7 • 6 • 5 • 4 • 3 • 2 • 1 3 • 2 • 1 Try This: Example 2B There are 7 soup cans in the pantry. B. If the shelf has only enough room for 4 cans, find the number of ways 4 of the 7 cans can be arranged. 7P4 = = 7 • 6 • 5 • 4 = 840 There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways.

  17. A combination is a selection of things in any order.

  18. 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.

  19. 60 10 In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is = 10. 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.

  20. 10! 2!(10 – 2)! = 10 possible books 10! 2!8! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 2 books chosen at a time (2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) Additional Example 3A: Finding 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. 10C2 = = = 45 There are 45 combinations. This means that Mary can buy 45 different pairs of books.

  21. 10! 7!(10 – 7)! = 10 possible books 10! = 7!3! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 7 books chosen at a time (7 • 6 • 5 • 4 • 3 • 2 • 1)(3 • 2 • 1) Additional Example 3B: Finding Combinations B. If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy. 10C7 = = 120 There are 120 combinations. This means that Mary can buy 120 different sets of 7 books.

  22. 12! 4!(12 – 4)! = 12 possible DVDs 12! 4!8! 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 4 DVDs chosen at a time (4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) Try This: Example 3A 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. 12C4 = = 495

  23. Try This: Example 3A Continued There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs.

  24. 12! 11!(12 – 11)! = 12 possible DVDs 12! 11!1! 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 11 DVDs chosen at a time (11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)(1) Try This: Example 3B B. If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy. 12C11 = = 12

  25. Try This: Example 3B Continued There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs.

  26. 9! 5! 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? 4. A group of 12 people are forming a committee. How many different 4-person committees can be formed? 362,880 3024 40,320 495

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