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§12.6 Permutations & Combinations

§12.6 Permutations & Combinations. 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

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§12.6 Permutations & Combinations

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  1. §12.6 Permutations & Combinations

  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 12 10,000

  3. Reading Math Read 5! as “five factorial.” Factorials 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! = 5 • 4 • 3 • 2 • 1

  4. 8 •7 • 6 • 5 • 4 • 3 • 2 • 1 6 • 5 • 4 • 3 • 2 • 1 Example 1: 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

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

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

  7. first letter second letter third letter ? ? ? 5 • 4 • 3 • 2 • 1 5! 2! = 2 • 1 Permutations 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 =

  8. Permutations

  9. 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 Example 2: 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.

  10. 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 = Example 2: 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. 6P3 = 6 • 5 • 4 = 120 There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways.

  11. Combinations A combination is a selection of things in any order. 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.

  12. 60 10 In the list of 60 permutations, each combination is repeated 6 times. The number of combinations is = 10. Combinations 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.

  13. Combinations

  14. 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) Example 3: 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.

  15. 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) Example 3: 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.

  16. Assignment: Pg. 766 1-39 Left

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