ACTIVITY 6.7 SELECTING AND REARRANGING THINGS 757 OBJECTIVES ACTIVITY 6.7 Selecting and Rearranging Things 1. Determine the number of permutations. 2. Determine the number of combinations. 3. Recognize patterns modeled by counting techniques. In working through the previous activities, you may have noticed there are many different varieties of counting problems. Sometimes a tree or list may be feasible to display all the possibilities. But usually the overwhelming number of possibilities makes constructing a list or tree impractical. In this activity you will explore two types of counting problems, permutations and combinations, both of which use the multiplication principle of counting. Permutations are arrangements of objects, from first to last, where the order in which the objects are selected is most important. Example 1 You havefivetextbooks and wish to arrange them on your bookshelf in the usual way. In how many ways can this be done? Think of the five positions on the shelf, left to right. Consider how many ways the first space can be filled (5), then how many ways to fill the second space (4, since there are four books remaining), and so on. You could illustrate with a tree diagram if you wish, but a simple application of the multiplication principle will give you 5 4 3 2 1 = 120. So there are 120 different arrangements or orderings of your five books. Equivalently, you say there are 120 permutations of any five objects. 1. Suppose you have seven different sweatshirts, and wish to wear a different one each day of the week. In how many ways can this be done? 2. There are 26 letters in the standard English alphabet. If you consider using all the letters to create a 26-letter word, how many different words are possible? (You will notice this is a very large number, so you will need scientific notation to give an approximate answer.)
758 CHAPTER 6 PROBABILITY MODELS Factorial Notation The product of any positive integer with all smaller positive integers down to one has a special name and notation. For example, Five factorial is written as 51 =5.4. 3.2-1 = 120. Appendix 3. Use the factorial notation on your calculator (see Appendix A) to determine 12! Note: 0! is defined to be equal to one. (Check this on your calculator.) You can think of it as the number of ways to arrange zero things. Sometimes you are interested in arrangements of only some of the objects from a larger collection. Words that are 26 letters long are a bit unrealistic, but using only five different letters might make more sense. Example 2 Suppose exactlyfivedifferent letters are used to make a word (in this case a word does not have to be one found in a dictionary, but simply be a sequence of five letters). How many different words can be made? As with the bookshelf problem, imagine five spaces, to hold each of the five letters. But now, there are 26 choices for the first space, 25 for the second, and so on. 26 25 24 23 22 = 7,893,600 different words are possible. Notice that since there are only five letters, this is not a full factorial. These are still called permutations. There are 7,893,600 permutations of 26 objects, taken five at a time. Permutations of n objects taken r at a time = the first r factors of n! Notation: np r = n(n - 1)(«- 2) (n - r + 1) = (n-r)! [A]*- A*,,,**, 4. Redo Example 2 using the permutations formula above. Then, verify your answer by using the permutation feature on your calculator (see Appendix A). 5. The governor is visiting your school and you need to assign seating for 14 students. Unfortunately there are only six seats in the front row. In how many different ways can you select and arrange the six lucky students? Combinations Sometimes the order in which selections are made is not important. For example, in Problem 5, if the arrangement of the six chosen students is not important, you would simply want to count how many ways a group of six students could be chosen for the first row. For this revised situation you are counting the number of combinations of 14 objects taken six at a time.
ACTIVITY 6.7 SELECTING AND REARRANGING THINGS 759 Combinations are collections of objects selected from a larger collection, where the order of selection is not important. The number of ways to select r objects from a collection of n objects is called the number of combinations of n objects taken r at a time. 6. In Problem 5, for each possible group of six students, how many seating arrangements are possible? 7. If you could list all 2,162,160 permutations in Problem 5, you could also group them according to the actual six students selected. Each group would have the number you determined in Problem 6. Divide accordingly to determine the number of ways to select the six students. P Combinations of n objects taken r at a time: nc r =!L rp- r «' (n-r)lr\ 8. Apply the combinations formula above to verify your answer to Problem 7. 9. From a student body of 230 students, how many different committees of four students are possible? (Hint: The order in which the four students are selected is not important.) An interesting symmetrical property of combinations appears when you consider not only the objects selected from a collection, but also those left behind. Does it make sense that the number of ways to select 3 objects from a collection of 8 objects is the same as counting the number of ways to leave 5 behind? 10. Verify the above statement by computing 8C 3 and 8C 5. Many times the most important part of the problem is determining whether or not the order of the selections is significant. In Exercise 5 of Activity 6.2, involving ordering a triple scoop ice cream cone, some people might consider the order of the scoops important. Others might not care how the three flavors are placed on the cone. 11. In each situation, determine whether you are asked to determine the number of permutations or combinations. Then do the calculation. a. How many ways are there to pick a starting five from a basketball team of twelve members?
760 CHAPTER 6 PROBABILITY MODELS b. How many ways are there to distribute nine different books among 15 children if no child gets more than one book? c. How many ways are there to pick a subset of four different letters from the 26-letter alphabet? d. How many ways can the three offices of chairman, vice chairman, and secretary be filled from a club with 25 members? Applications Many variations on these basic counting techniques arise in the study of probability and computer science. Here are a couple of examples, with problems for you to try. «Example 3 How many different ten-digit binary sequences (only 0s and Is) are there with exactly seven zeros and three ones? While the order of the binary digits is important, this counting problem is most easily solved by determining the number of combinations of ten digits taken seven at a time. Think of the ten positions for the digits, and ask: How many ways can you select seven positions for the zeros? 10 9 8 7 6 5 4 3 2 1.10-9 low ~ (7 6 5 4 3 2 1) (3 2 1) 3 2 = 120. Or equivalently, how many ways can you select three positions for the ones? 10 9 8 7 6 5 4 3 2 1 10-9 IOW C, = (3 2 1) (7 6 5 4 3 2 1) 3 2 = 120. 12. If a coin is flipped twenty times, how many different ways are there to get exactly five heads? 13. What is the probability that a coin flipped twenty times will come up heads exactly five times?
ACTIVITY 6.7 SELECTING AND REARRANGING THINGS 761 Example 4 How many ways are there to arrange the letters of the word SYSTEMS? T S S Y MS E The repeated S makes this more difficult than a simple permutation problem. If you imagined the letters on tiles (like the game Scrabble), with the S tiles labeled with numbers: S, S 2 * S3, so you can tell them apart, then there would be simply 7! ways to arrange the seven tiles. But without the numbers, or tiles, the S's all look the same, so you would be way over-counting the different words. Each unique word will appear 3! times, the number of ways the three 5's can be ordered. So if you divide out this multiplicity, you will have an accurate count: 7' = 7 6 5 4 = 840 truly unique words. 14. How many different words can be formed by rearranging all the letters of the word MIRROR"! SUMMARY ACTIVITY 6.7 1. Permutations are arrangements of objects, from first to last, where the order in which the objects are selected is most important. 2. Factorial Notation: n\ ~ n(n - l)(n - 2) 3 2 1 3. Permutations of n objects taken r at a time = the first r factors of n! Notation: np r = n(n - l)(n - 2) n\ (n - r + 1) = (n - r)! 4. Combinations are collections of objects selected from a larger collection, where the order of selection is not important. r n ' 5. Combinations of n objects taken r at a time = C r = "- = 7 '-r-, r\