10 -9 Permutations and Combinations Warm Up Problem of the Day Lesson Presentation Course 3
10 -9 Permutations and Combinations Warm Up Find the number of possible outcomes. 1. bagels: plain, egg, wheat, onion meat: turkey, ham, roast beef, tuna 16 2. eggs: scrambled, over easy, hard boiled meat: sausage patty, sausage link, bacon, ham 12 3. How many different 4–digit phone extensions are possible? 10, 000 Course 3
10 -9 Permutations and Combinations 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 and Combinations Learn to find permutations and combinations. Course 3
10 -9 Permutations and Combinations Insert Lesson Title Here Vocabulary factorial permutation combination 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 Reading Math Read 5! as “five factorial. ” Course 3
10 -9 Permutations and Combinations Additional Example 1: Evaluating Expressions Containing Factorials Evaluate each expression. A. 9! 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 362, 880 B. 8! 6! 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 8 • 7 = 56 Course 3 Write out each factorial and simplify. Multiply remaining factors.
10 -9 Permutations and Combinations Additional Example 1: Evaluating Expressions Containing Factorials 10! C. (9 – 2)! 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 Course 3
10 -9 Permutations and Combinations Check It Out: Example 1 Evaluate each expression. A. 10! 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 3, 628, 800 B. 7! 5! 7 • 6 • 5 • 4 • 3 • 2 • 1 7 • 6 = 42 Course 3 Write out each factorial and simplify. Multiply remaining factors.
10 -9 Permutations and Combinations Check It Out: Example 1 9! C. (8 – 2)! 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 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 choices • 2 choices • 1 choice The product can be written as a factorial. 3 • 2 • 1 = 3! = 6 Course 3
10 -9 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 choices 4 choices 3 choices = 60 permutations Notice that the product can be written as a quotient of factorials. 5! 60 = 5 • 4 • 3 • 2 • 1 = 2! 2 • 1 Course 3
10 -9 Permutations and Combinations Remember! By definition, 0! = 1. Course 3
10 -9 Permutations and Combinations Additional Example 2 A: Finding Permutations Jim has 6 different books. Find the number of orders in which the 6 books can be arranged on a shelf. The number of books is 6. 6! = 6 • 5 • 4 • 3 • 2 • 1 = 720 P 6 = 6 (6 – 6)! 0! 1 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. Course 3
10 -9 Permutations and Combinations Additional Example 2 B: 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. 6! = 6 • 5 • 4 • 3 • 2 • 1 = 6 • 5 • 4 P 3 = 6 (6 – 3)! 3! 3 • 2 • 1 The books are = 120 arranged 3 at a time. There are 120 permutations. This means that 3 of the 6 books can be arranged in 120 ways. Course 3
10 -9 Permutations and Combinations Check It Out: Example 2 A 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. The number of cans is 7. 7! = 7 • 6 • 5 • 4 • 3 • 2 • 1 = 5040 P 7 = 7 (7 – 7)! 0! 1 The cans are arranged 7 at a time. There are 5040 orders in which to arrange 7 soup cans. Course 3
10 -9 Permutations and Combinations Check It Out: Example 2 B 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! = 7 • 6 • 5 • 4 • 3 • 2 • 1 P 4 = 7 (7 – 4)! 3! 3 • 2 • 1 The cans are arranged = 7 • 6 • 5 • 4 = 840 4 at a time. There are 840 permutations. This means that the 7 cans can be arranged in the 4 spaces in 840 ways. Course 3
10 -9 Permutations and Combinations A combination is a selection of things in any order. Course 3
10 -9 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. Course 3
10 -9 Permutations and Combinations ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE ACB ADB AEB ADC AED BDC BED CED BAC BAD BAE CAD CAE DAE CBD CBE DCE BCA BDA BEA CDA CEA DBC CEB DEC CAB DAB EAB DAC EAD DCB EBC EBD ECD CBA DBA EBA DCA EDA DBC ECB 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 60 = 10. 6 Course 3
10 -9 Permutations and Combinations Course 3
10 -9 Permutations and Combinations Additional Example 3 A: Finding Combinations Mary wants to join a book club that offers a choice of 10 new books each month. If Mary wants to buy 2 books, find the number of different pairs she can buy. 10 possible books 10 C 2 = 10! 2!(10 – 2)! 2!8! 2 books chosen at a time = 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 45 (2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) There are 45 combinations. This means that Mary can buy 45 different pairs of books. Course 3
10 -9 Permutations and Combinations Additional Example 3 B: Finding Combinations If Mary wants to buy 7 books, find the number of different sets of 7 books she can buy. 10 possible books 10 C 7 = 10! 7!(10 – 7)! 7!3! 7 books chosen at a time = 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 120 (7 • 6 • 5 • 4 • 3 • 2 • 1)(3 • 2 • 1) There are 120 combinations. This means that Mary can buy 120 different sets of 7 books. Course 3
10 -9 Permutations and Combinations Check It Out: Example 3 A Harry wants to join a DVD club that offers a choice of 12 new DVDs each month. If Harry wants to buy 4 DVDs, find the number of different sets he can buy. 12 possible DVDs 12 C 4 = 12! 4!(12 – 4)! 4!8! 4 DVDs chosen at a time = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (4 • 3 • 2 • 1)(8 • 7 • 6 • 5 • 4 • 3 • 2 • 1) = 495 Course 3
10 -9 Permutations and Combinations Check It Out: Example 3 A Continued There are 495 combinations. This means that Harry can buy 495 different sets of 4 DVDs. Course 3
10 -9 Permutations and Combinations Check It Out: Example 3 B If Harry wants to buy 11 DVDs, find the number of different sets of 11 DVDs he can buy. 12 possible DVDs 12 C 11 = 12! 11!(12 – 11)! 11!1! 11 DVDs chosen at a time = 12 • 11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 (11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1)(1) = 12 Course 3
10 -9 Permutations and Combinations Check It Out: Example 3 B Continued There are 12 combinations. This means that Harry can buy 12 different sets of 11 DVDs. Course 3
10 -9 Permutations and Combinations Insert Lesson Title Here Lesson Quiz Evaluate each expression. 1. 9! 362, 880 2. 9! 3024 5! 3. There are 8 hot air balloons in a race. In how many possible orders can all 8 hot air balloons finish the race? 40, 320 4. A group of 12 people are forming a committee. How many different 4 -person committees can be 495 formed? Course 3
Скачать презентацию 10 -9 Permutations and Combinations Warm Up Problem
ea44058561abd8479162dd69c271ac3d.ppt