Betsy McCall Courses -- Math Notes Basic Counting Principles III -- No Repetition, Order Matters -- Permutations


homepage courses fall2004 spring2005 summer 2005 fall 2005 spring 2006 summer 2006 fall 2006 spring 2007 summer 2007 fall 2007 winter 2008 spring 2008 summer 2008 fall 2008 mathnotes resources previous -- view my teaching portfolio here		Math Notes Counting Methods, No Repetition, Order Matters: Permutations Introduction Permutations involve all the possible arrangements (where order matters with no repetitions) of a fixed set of objects. Consider the following set of four numbers {5, 6, 7, 8}. All the possible permutations of this set of numbers are: 5678 5687 5867 5876 6578 6587 6578 5786 5768 8756 8765 6758 6785 6857 6875 7568 7586 7658 7685 7865 7856 8567 8576 8657 8675 There are 24 possible permutations of these four objects. As the number of objects in the set grows, the number of possible permutations grows much more rapidly. If our set of objects has six elements in it, the number of possible permutations is 720. Sometimes we wish to select fewer elements in the set than are present. For the same set above, if we choose two elements, the possible combinations are: 56 57 58 65 67 68 75 76 78 85 86 87 There are twelve possible permutations. There is a formula we can use to find the number of possible permutations. It relates the total number of items in the set, n, with the total number of elements to be chosen, r. The formula is sometimes referred to with the label nPr or P(n, r), and is given by the relation n!/(n-r)! What this essentially tells us, is that the number of permutations is calculated by multiplying the number n by (n-1) by (n-2) and so on until you have the number r of terms you wish to count. [Remember that factorials, 4! for instance, just means 4321.] For instance, to calculate the number of permutations possible from a set of four elements when one chooses two elements, one multiplies 43 = 12. Equivalently we can use our expression above, 4!/(4-2)! = 4!/2! = 4321/(21) = 43 = 12. If one wishes to calculate the number of possible permutations from a set of four elements as we did above by listing, when one is choosing all four elements, one multiplies 4321 = 24. Equivalently, we can use our expression above 4!/(4-4)! = 4!/0! Remember that 0! is defined to be 1 so this is just 4! = 4321 = 24. Worked Examples* A. What are the number of possible permutations of a set of ten elements (using all ten)? We can use our formula: n = 10 since that's the size of our set, and r = 10 since we are using all the elements. this gives us 10!/(10-10)! = 10!/0! = 10! = 3,628,800. B. How many possible permutations are there of a two elements from a set of three elements, such as abc? We can use our formula by determining the values for n and r. The size of our set is 3, and we want to use only two elements at a time. Therefore, 3!/(3-2)! = 3!/1! = 3! = 321 = 6. C. How many possible permutations are there of four elements from a set of ten elements, such as 0123456789? Here n = 10 and r = 3. 10!/(10-3)! = 10!/7! = 10987654321/(7654321) = 1098 = 720, which is exactly what we'd expect to get using the fundamental counting principle. Problem Solving Tips* There are some things to look for in problems to tell if you are supposed to use permutations. Does order matter? If you answered yes, permutations may work. Is repetition allowed? If you answered yes, then permutations won't work. Sometimes your problem will say so explicitly, but usually it won't. If you are choosing balls in a large state lottery, each number can only be chosen once, and you win more if you get the order correct as well, so permutations would be appropriate. Whenever you are ordering people since people can't be in two places at once, permutations are appropriate if order matters. For instance anytime you are assigning officers or first and second place. If your unsure, put yourself into the problem... would you care if you were chosen first or second? If the answer is yes, then use permutations. Do you have a number of objects with the same label? Like blue balls or something? Then order is not really relevant, so you have to use something else. If you are using a TI-83 calculator, you can find factorial and permutations formulae preprogrammed into your calculator. Hit the Math menu button. You have four menu options. Scroll over to the PRB menu. Factorial is command 1 under this menu and permutations as nPr is command 2. For both of these type a number first, then hit the factorial or permutation command. For factorial hit enter; for permutations, type the second number and then hit enter. Additional Problems with Answers 1. How many different arrangements (permutations) are there of the digits 01234? 2. What is the value of P(7, 5)? 3. Suppose that sixteen people show up at a theatre but inside the main doors there is only room for a queue of ten people between the doors and the ticket window. Find the number of such queues of ten people. 4. You are choosing a four-character password involving both letters and numbers (case of the letter is not important). Determine the number of possible four-character passwords that can be used without repeating any characters. 5. You are selecting five cards from a standard deck of playing cards (there are 52 cards in a standard deck). How many different ways can you choose those five cards? 6. In how many ways can 8 CD’s be arranged on a shelf? 7. If a softball league has 12 teams, how many different end of the season rankings are possible? (Assume no ties). 8. In how many ways can the scrabble club of 20 members select a president, vice president and treasury, assuming that the same person cannot hold more than one office. 9. A key pad lock has 10 different digits, and a sequence of 5 different digits must be selected for the lock to open. How many key pad combinations are possible (suppose the keys remain pressed until the entire code is entered so that digits can't be used again)? Challenge Problem: 10. How many 5-digit even numbers can you make from {1245789}? Answers: 1. 120 2. 2520 3. 2.9 x 10¹⁰ 4. 1,413,720 5. 311,875,200 6. 40,320 7. 479,001,600 8. 6840 9. 30,240 10. 2520 Links to Outside Sources Permutation -- from MathWorld College Algebra Tutorial on Permutations Permutations Activity Links to Supporting Topics Counting Methods I -- With Repetition, Order Matters: Fundamental Counting Principle Counting Methods II -- With Repetition, Order Doesn't Matter Counting Methods IV -- No Repetition, Order Doesn't Matter: Combinations Math Forum: Ask Dr. Math FAQ: Permutations and Combinations 7.6 - Counting Principles Counting: Permutations and Combinations Even and Odd Permutations Permutations in Abstract Algebra History Group theory history of restricted permutations History of ringing Journal of Online Mathematics and its Applications \| Parity Theorem for Permutations Info on Permutations

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