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9-6

9-6. 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 filling: turkey, ham, roast beef, tuna, chicken, cheese, sourcream

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9-6

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  1. 9-6 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 filling: turkey, ham, roast beef, tuna, chicken, cheese, sourcream 2. eggs: scrambled, over easy, hard boiled, raw meat: sausage patty, sausage link, bacon, ham, steak, pork 3. How many different 3–digit phone extensions are possible?

  3. Learn to find combinations.

  4. Vocabulary combination

  5. 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

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

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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

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

  13. 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

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

  15. 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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