{"id":2466,"date":"2017-03-31T21:47:54","date_gmt":"2017-03-31T21:47:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=2466"},"modified":"2019-05-30T17:01:06","modified_gmt":"2019-05-30T17:01:06","slug":"counting","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/counting\/","title":{"raw":"Counting","rendered":"Counting"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Compute a conditional probability for an event<\/li>\r\n \t<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\r\n \t<li>Calculate the expected value of an event<\/li>\r\n<\/ul>\r\n<\/div>\r\nCounting? You already know how to count or you wouldn't be taking a college-level math class, right? Well yes, but what we'll really be investigating here are ways of counting <em>efficiently<\/em>. When we get to the probability situations a bit later in this chapter we will need to count some <em>very<\/em> large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don't want to do this.\r\n<h3>Basic Counting<\/h3>\r\nWe will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?\r\n\r\n<strong>Solution 1<\/strong>: One way to solve this problem would be to systematically list each possible meal:\r\n\r\nsoup + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + quiche\r\n\r\nsoup + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + hamburger\r\n\r\nsalad + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + quiche \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + fajita\r\n\r\nsalad + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + sandwich\r\n\r\nbreadsticks + quiche \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + pizza\r\n\r\nAssuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.\r\n\r\n<strong>Solution 2<\/strong>: Another way to solve this problem would be to list all the possibilities in a table:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>hamburger<\/td>\r\n<td>sandwich<\/td>\r\n<td>quiche<\/td>\r\n<td>fajita<\/td>\r\n<td>pizza<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>soup<\/td>\r\n<td>soup+burger<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>salad<\/td>\r\n<td>salad+burger<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>bread<\/td>\r\n<td>etc<\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIn each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn't really care <em>what<\/em> the possible meals are, only <em>how many<\/em> possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It's always good when you solve a problem two different ways and get the same answer!)\r\n\r\n<strong>Solution 3<\/strong>: We already have two perfectly good solutions. Why do we need a third?\u00a0 The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we've used the rows of the table for the appetizers and the columns for the main courses\u2014where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn't terribly easy, we need a better way in case we have three categories to choose form instead of just two.\r\n\r\nSo, back to the problem in the example.\u00a0 What else can we do?\u00a0 Let's draw a <strong>tree diagram<\/strong>:\r\n\r\n<img class=\"aligncenter size-full wp-image-341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11181317\/treediagram.png\" alt=\"Tree diagram of above table\" width=\"375\" height=\"225\" \/>\r\n\r\nThis is called a \"tree\" diagram because at each stage we branch out, like the branches on a tree.\u00a0 In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer).\u00a0 We count the number of branches at the final level and get (surprise, surprise!) 15.\r\n\r\nIf we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.\r\n\r\n<\/div>\r\nOK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?\r\n\r\nLet's go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the <em>basic counting rule<\/em>:\r\n<div class=\"textbox\">\r\n<h3>Basic Counting Rule<\/h3>\r\nIf we are asked to choose one item from each of two separate categories where there are <em>m<\/em> items in the first category and <em>n<\/em> items in the second category, then the total number of available choices is <em>m <\/em><strong>\u00b7 <\/strong><em>n<\/em>.\r\n\r\nThis is sometimes called the multiplication rule for probabilities.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nThere are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?\r\n[reveal-answer q=\"16976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16976\"]There are 21 choices from the first category and 18 for the second, so there are 21\u00a0<strong>\u00b7<\/strong>\u00a018\u00a0=\u00a0378 possibilities.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nSuppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?\r\n[reveal-answer q=\"13816\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"13816\"]\r\n\r\nThere are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are 3 <strong>\u00b7<\/strong> 5 <strong>\u00b7<\/strong> 2 = 30 possibilities.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7159&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA quiz consists of 3 true-or-false questions.\u00a0 In how many ways can a student answer the quiz?\r\n[reveal-answer q=\"169025\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"169025\"]There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 = 8 different ways.\u00a0 Recall that another way to write 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 is 2<sup>3<\/sup>, which is much more compact.[\/hidden-answer]\r\n\r\n<\/div>\r\nBasic counting examples from this section are described in the following video.\r\n\r\nhttps:\/\/youtu.be\/fROqcu-ekkw\r\n<h2>Permutations<\/h2>\r\nIn this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means.\u00a0 Let's start with a couple examples.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nHow many different ways can the letters of the word MATH be rearranged to form a four-letter code word?\r\n[reveal-answer q=\"614648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"614648\"]This problem is a bit different.\u00a0 Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the <em>same<\/em> category (the category is: the letters of the word MATH) and each time we choose an item we <em>do not replace<\/em> it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T).\u00a0 Thus there are 4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 24 ways to spell a code worth with the letters MATH.[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this example, we needed to calculate <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1. This calculation shows up often in mathematics, and is called the <strong>factorial<\/strong>, and is notated <em>n<\/em>!\r\n<div class=\"textbox\">\r\n<h3>Factorial<\/h3>\r\n<p style=\"text-align: center;\"><em>n<\/em>! = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7188&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nHow many ways can five different door prizes be distributed among five people?\r\n[reveal-answer q=\"530386\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530386\"]There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 <strong>\u00b7 <\/strong>4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 120 ways.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7191&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\nNow we will consider some slightly different examples.\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nA charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10.\u00a0 Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?\r\n[reveal-answer q=\"75787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"75787\"]Using the Basic Counting Rule, there are 25 choices for the person who receives the $100 certificate, 24 remaining choices for the $25 certificate and 23 choices for the $10 certificate, so there are 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 = 13,800 ways in which the prizes can be awarded.[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?\r\n[reveal-answer q=\"68166\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68166\"]\r\n\r\nUsing the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 <strong>\u00b7<\/strong> 7 <strong>\u00b7<\/strong> 6 = 336 ways the three medals can be awarded to the 8 runners.\r\n\r\nNote that in these preceding examples, the gift certificates and the Olympic medals were awarded <em>without replacement<\/em>; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second).\u00a0 Contrast this with the situation of a multiple choice test, where there might be five possible answers\u00a0\u2014 A, B, C, D or E \u2014 for each question on the test.\r\n\r\nNote also that <em>the order of selection was important<\/em> in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case the order of selection is <em>not<\/em> important since each of the three people receive the same prize.\u00a0 Situations where the order is <em>not<\/em> important will be discussed in the next section.)\r\n\r\n[\/hidden-answer]\r\n\r\nFactorial examples are worked in this video.\r\n\r\nhttps:\/\/youtu.be\/9maoGi5fd_M\r\n\r\n<\/div>\r\nWe can generalize the situation in the two examples above to any problem <em>without replacement<\/em> where the <em>order of selection is important<\/em>. If we are arranging in order <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by\r\n<p style=\"text-align: center;\"><em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\r\nIf you don't see why (<em>n <\/em>\u2014 <em>r <\/em>+ 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 to get 13,800. In this case <em>n<\/em> = 25 and <em>r<\/em> = 3, so <em>n <\/em>\u2014 <em>r <\/em>+ 1 = 25 \u2014 3 + 1 = 23, which is exactly the right number for the final factor.\r\n\r\nNow, why would we want to use this complicated formula when it's actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won't actually use this formula all that often; we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection is important<\/em>. In this situation we write:\r\n<div class=\"textbox\">\r\n<h3>Permutations<\/h3>\r\n<p style=\"text-align: center;\"><em>nPr<\/em> = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\r\nWe say that there are <em>nPr<\/em> <strong>permutations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> when <em>order matters<\/em>.\r\n\r\nIt turns out that we can express this result more simply using factorials.\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{P}_{r}}=\\frac{n!}{(n-r)!}[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nI have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?\r\n[reveal-answer q=\"366612\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"366612\"]Since we are choosing 4 paintings out of 9 <em>without replacement<\/em> where the <em>order of selection is important<\/em> there are 9<em>P<\/em>4 = 9 \u00b7 8 \u00b7 7 \u00b7 6 = 3,024 permutations.[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nHow many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?\r\n[reveal-answer q=\"889810\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889810\"]We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is 16<em>P<\/em>4 = 16 \u00b7 15 \u00b7 14 \u00b7 13 = 43,680.[\/hidden-answer]\r\n\r\nView this video to see more about the permutations examples.\r\n\r\nhttps:\/\/youtu.be\/xlyX2UJMJQI\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nHow many 5 character passwords can be made using the letters A through Z\r\n<ul>\r\n \t<li>if repeats are allowed<\/li>\r\n \t<li>if no repeats are allowed<\/li>\r\n<\/ul>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7193&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Combinations<\/h2>\r\nIn the previous section we considered the situation where we chose <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection was important<\/em>. We now consider a similar situation in which the order of selection is <em>not<\/em> important.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?\r\n[reveal-answer q=\"942898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942898\"]\r\n\r\nUsing the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 \u00b7 24 \u00b7 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn't really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:\r\n\r\nABC\u00a0\u00a0\u00a0\u00a0 ACB\u00a0\u00a0\u00a0\u00a0 BAC\u00a0\u00a0\u00a0\u00a0 BCA\u00a0\u00a0\u00a0\u00a0 CAB\u00a0\u00a0\u00a0\u00a0 CBA\r\n\r\nHow can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 \u00b7 2 \u00b7 1 = 6 ways to do this; we didn't really need to list them all. We can just use permutations!\r\n\r\nSo, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted <em>six times<\/em>. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800\/6 = 2300.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nWe can generalize the situation in this example above to any problem of choosing a collection of items <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>. If we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by [latex]\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex], and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>.\r\n<div class=\"textbox\">\r\n<h3>Combinations<\/h3>\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex]<\/p>\r\nWe say that there are <em>nCr<\/em> <strong>combinations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> where <em>order doesn\u2019t matter<\/em>.\r\n\r\nWe can also write the combinations formula in terms of factorials:\r\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{n!}{(n-r)!r!}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?\r\n[reveal-answer q=\"125383\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"125383\"]Since we are choosing 4 people out of 35 <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em> there are [latex]{}_{35}{{C}_{4}}=\\frac{35\\cdot34\\cdot33\\cdot32}{4\\cdot3\\cdot2\\cdot1}[\/latex] = 52,360 combinations.[\/hidden-answer]\r\n\r\nView the following for more explanation of the combinations examples.\r\n\r\nhttps:\/\/youtu.be\/W8kd4YosbzE\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?\r\n\r\n<\/div>\r\nIn the preceding Try It problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee.\u00a0 (The same of course would be true if the Democrats were in control.) So let's consider the problem again, in a slightly more complicated form:\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?\r\n[reveal-answer q=\"205320\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205320\"]\r\n\r\nIn this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats.\u00a0 There are 15<em>C<\/em>10 = 3003 ways to choose the 10 Republicans and 14<em>C<\/em>9 = 2002 ways to choose the 9 Democrats. But now what?\u00a0 How do we finish the problem?\r\n\r\nSuppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and \u00a0all of the possible 9-member Democratic groups on 2002 slips of blue paper.\u00a0 How many ways can we choose one red slip and one blue slip?\u00a0 This is a job for the Basic Counting Rule!\u00a0 We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.\r\n\r\nThere must be 3003 \u00b7 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.\r\n\r\n[\/hidden-answer]\r\n\r\nThis example is worked through below.\r\n\r\nhttps:\/\/youtu.be\/Xqc2sdYN7xo\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Compute a conditional probability for an event<\/li>\n<li>Use Baye\u2019s theorem to compute a conditional probability<\/li>\n<li>Calculate the expected value of an event<\/li>\n<\/ul>\n<\/div>\n<p>Counting? You already know how to count or you wouldn&#8217;t be taking a college-level math class, right? Well yes, but what we&#8217;ll really be investigating here are ways of counting <em>efficiently<\/em>. When we get to the probability situations a bit later in this chapter we will need to count some <em>very<\/em> large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don&#8217;t want to do this.<\/p>\n<h3>Basic Counting<\/h3>\n<p>We will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?<\/p>\n<p><strong>Solution 1<\/strong>: One way to solve this problem would be to systematically list each possible meal:<\/p>\n<p>soup + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + quiche<\/p>\n<p>soup + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 soup + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + hamburger<\/p>\n<p>salad + sandwich \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + quiche \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 salad + fajita<\/p>\n<p>salad + pizza \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + hamburger \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + sandwich<\/p>\n<p>breadsticks + quiche \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + fajita \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 breadsticks + pizza<\/p>\n<p>Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.<\/p>\n<p><strong>Solution 2<\/strong>: Another way to solve this problem would be to list all the possibilities in a table:<\/p>\n<table>\n<tbody>\n<tr>\n<td><\/td>\n<td>hamburger<\/td>\n<td>sandwich<\/td>\n<td>quiche<\/td>\n<td>fajita<\/td>\n<td>pizza<\/td>\n<\/tr>\n<tr>\n<td>soup<\/td>\n<td>soup+burger<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>salad<\/td>\n<td>salad+burger<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>bread<\/td>\n<td>etc<\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn&#8217;t really care <em>what<\/em> the possible meals are, only <em>how many<\/em> possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It&#8217;s always good when you solve a problem two different ways and get the same answer!)<\/p>\n<p><strong>Solution 3<\/strong>: We already have two perfectly good solutions. Why do we need a third?\u00a0 The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we&#8217;ve used the rows of the table for the appetizers and the columns for the main courses\u2014where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn&#8217;t terribly easy, we need a better way in case we have three categories to choose form instead of just two.<\/p>\n<p>So, back to the problem in the example.\u00a0 What else can we do?\u00a0 Let&#8217;s draw a <strong>tree diagram<\/strong>:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-341\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11181317\/treediagram.png\" alt=\"Tree diagram of above table\" width=\"375\" height=\"225\" \/><\/p>\n<p>This is called a &#8220;tree&#8221; diagram because at each stage we branch out, like the branches on a tree.\u00a0 In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer).\u00a0 We count the number of branches at the final level and get (surprise, surprise!) 15.<\/p>\n<p>If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.<\/p>\n<\/div>\n<p>OK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?<\/p>\n<p>Let&#8217;s go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the <em>basic counting rule<\/em>:<\/p>\n<div class=\"textbox\">\n<h3>Basic Counting Rule<\/h3>\n<p>If we are asked to choose one item from each of two separate categories where there are <em>m<\/em> items in the first category and <em>n<\/em> items in the second category, then the total number of available choices is <em>m <\/em><strong>\u00b7 <\/strong><em>n<\/em>.<\/p>\n<p>This is sometimes called the multiplication rule for probabilities.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>There are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16976\">Show Solution<\/span><\/p>\n<div id=\"q16976\" class=\"hidden-answer\" style=\"display: none\">There are 21 choices from the first category and 18 for the second, so there are 21\u00a0<strong>\u00b7<\/strong>\u00a018\u00a0=\u00a0378 possibilities.<\/div>\n<\/div>\n<\/div>\n<p>The Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q13816\">Show Solution<\/span><\/p>\n<div id=\"q13816\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 3 choices for an appetizer, 5 for the main course and 2 for dessert, so there are 3 <strong>\u00b7<\/strong> 5 <strong>\u00b7<\/strong> 2 = 30 possibilities.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7159&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A quiz consists of 3 true-or-false questions.\u00a0 In how many ways can a student answer the quiz?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q169025\">Show Solution<\/span><\/p>\n<div id=\"q169025\" class=\"hidden-answer\" style=\"display: none\">There are 3 questions. Each question has 2 possible answers (true or false), so the quiz may be answered in 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 = 8 different ways.\u00a0 Recall that another way to write 2 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 2 is 2<sup>3<\/sup>, which is much more compact.<\/div>\n<\/div>\n<\/div>\n<p>Basic counting examples from this section are described in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Basic counting\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fROqcu-ekkw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Permutations<\/h2>\n<p>In this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means.\u00a0 Let&#8217;s start with a couple examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>How many different ways can the letters of the word MATH be rearranged to form a four-letter code word?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q614648\">Show Solution<\/span><\/p>\n<div id=\"q614648\" class=\"hidden-answer\" style=\"display: none\">This problem is a bit different.\u00a0 Instead of choosing one item from each of several different categories, we are repeatedly choosing items from the <em>same<\/em> category (the category is: the letters of the word MATH) and each time we choose an item we <em>do not replace<\/em> it, so there is one fewer choice at the next stage: we have 4 choices for the first letter (say we choose A), then 3 choices for the second (M, T and H; say we choose H), then 2 choices for the next letter (M and T; say we choose M) and only one choice at the last stage (T).\u00a0 Thus there are 4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 24 ways to spell a code worth with the letters MATH.<\/div>\n<\/div>\n<\/div>\n<p>In this example, we needed to calculate <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1. This calculation shows up often in mathematics, and is called the <strong>factorial<\/strong>, and is notated <em>n<\/em>!<\/p>\n<div class=\"textbox\">\n<h3>Factorial<\/h3>\n<p style=\"text-align: center;\"><em>n<\/em>! = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 3 \u00b7 2 \u00b7 1<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7188&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>How many ways can five different door prizes be distributed among five people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q530386\">Show Solution<\/span><\/p>\n<div id=\"q530386\" class=\"hidden-answer\" style=\"display: none\">There are 5 choices of prize for the first person, 4 choices for the second, and so on. The number of ways the prizes can be distributed will be 5! = 5 <strong>\u00b7 <\/strong>4 <strong>\u00b7 <\/strong>3 <strong>\u00b7<\/strong> 2 <strong>\u00b7<\/strong> 1 = 120 ways.<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7191&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>Now we will consider some slightly different examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10.\u00a0 Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q75787\">Show Solution<\/span><\/p>\n<div id=\"q75787\" class=\"hidden-answer\" style=\"display: none\">Using the Basic Counting Rule, there are 25 choices for the person who receives the $100 certificate, 24 remaining choices for the $25 certificate and 23 choices for the $10 certificate, so there are 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 = 13,800 ways in which the prizes can be awarded.<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68166\">Show Solution<\/span><\/p>\n<div id=\"q68166\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the Basic Counting Rule, there are 8 choices for the gold medal winner, 7 remaining choices for the silver, and 6 for the bronze, so there are 8 <strong>\u00b7<\/strong> 7 <strong>\u00b7<\/strong> 6 = 336 ways the three medals can be awarded to the 8 runners.<\/p>\n<p>Note that in these preceding examples, the gift certificates and the Olympic medals were awarded <em>without replacement<\/em>; that is, once we have chosen a winner of the first door prize or the gold medal, they are not eligible for the other prizes. Thus, at each succeeding stage of the solution there is one fewer choice (25, then 24, then 23 in the first example; 8, then 7, then 6 in the second).\u00a0 Contrast this with the situation of a multiple choice test, where there might be five possible answers\u00a0\u2014 A, B, C, D or E \u2014 for each question on the test.<\/p>\n<p>Note also that <em>the order of selection was important<\/em> in each example: for the three door prizes, being chosen first means that you receive substantially more money; in the Olympics example, coming in first means that you get the gold medal instead of the silver or bronze. In each case, if we had chosen the same three people in a different order there might have been a different person who received the $100 prize, or a different goldmedalist. (Contrast this with the situation where we might draw three names out of a hat to each receive a $10 gift certificate; in this case the order of selection is <em>not<\/em> important since each of the three people receive the same prize.\u00a0 Situations where the order is <em>not<\/em> important will be discussed in the next section.)<\/p>\n<\/div>\n<\/div>\n<p>Factorial examples are worked in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Counting using the factorial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9maoGi5fd_M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>We can generalize the situation in the two examples above to any problem <em>without replacement<\/em> where the <em>order of selection is important<\/em>. If we are arranging in order <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by<\/p>\n<p style=\"text-align: center;\"><em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\n<p>If you don&#8217;t see why (<em>n <\/em>\u2014 <em>r <\/em>+ 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 <strong>\u00b7<\/strong> 24 <strong>\u00b7<\/strong> 23 to get 13,800. In this case <em>n<\/em> = 25 and <em>r<\/em> = 3, so <em>n <\/em>\u2014 <em>r <\/em>+ 1 = 25 \u2014 3 + 1 = 23, which is exactly the right number for the final factor.<\/p>\n<p>Now, why would we want to use this complicated formula when it&#8217;s actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won&#8217;t actually use this formula all that often; we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection is important<\/em>. In this situation we write:<\/p>\n<div class=\"textbox\">\n<h3>Permutations<\/h3>\n<p style=\"text-align: center;\"><em>nPr<\/em> = <em>n<\/em>\u00a0\u00b7 (<em>n<\/em> \u2013 1) \u00b7 (<em>n<\/em> \u2013 2) \u00b7\u00b7\u00b7 (<em>n<\/em> \u2013 <em>r<\/em> + 1)<\/p>\n<p>We say that there are <em>nPr<\/em> <strong>permutations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> when <em>order matters<\/em>.<\/p>\n<p>It turns out that we can express this result more simply using factorials.<\/p>\n<p style=\"text-align: center;\">[latex]{}_{n}{{P}_{r}}=\\frac{n!}{(n-r)!}[\/latex]<\/p>\n<\/div>\n<p>In practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>I have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q366612\">Show Solution<\/span><\/p>\n<div id=\"q366612\" class=\"hidden-answer\" style=\"display: none\">Since we are choosing 4 paintings out of 9 <em>without replacement<\/em> where the <em>order of selection is important<\/em> there are 9<em>P<\/em>4 = 9 \u00b7 8 \u00b7 7 \u00b7 6 = 3,024 permutations.<\/div>\n<\/div>\n<hr \/>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889810\">Show Solution<\/span><\/p>\n<div id=\"q889810\" class=\"hidden-answer\" style=\"display: none\">We want to choose 4 people out of 16 without replacement and where the order of selection is important. So the answer is 16<em>P<\/em>4 = 16 \u00b7 15 \u00b7 14 \u00b7 13 = 43,680.<\/div>\n<\/div>\n<p>View this video to see more about the permutations examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Permutations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xlyX2UJMJQI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>How many 5 character passwords can be made using the letters A through Z<\/p>\n<ul>\n<li>if repeats are allowed<\/li>\n<li>if no repeats are allowed<\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=7193&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Combinations<\/h2>\n<p>In the previous section we considered the situation where we chose <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> and where the <em>order of selection was important<\/em>. We now consider a similar situation in which the order of selection is <em>not<\/em> important.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q942898\">Show Solution<\/span><\/p>\n<div id=\"q942898\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the Basic Counting Rule, there are 25 choices for the first person, 24 remaining choices for the second person and 23 for the third, so there are 25 \u00b7 24 \u00b7 23 = 13,800 ways to choose three people. Suppose for a moment that Abe is chosen first, Bea second and Cindy third; this is one of the 13,800 possible outcomes. Another way to award the prizes would be to choose Abe first, Cindy second and Bea third; this is another of the 13,800 possible outcomes. But either way Abe, Bea and Cindy each get $50, so it doesn&#8217;t really matter the order in which we select them. In how many different orders can Abe, Bea and Cindy be selected? It turns out there are 6:<\/p>\n<p>ABC\u00a0\u00a0\u00a0\u00a0 ACB\u00a0\u00a0\u00a0\u00a0 BAC\u00a0\u00a0\u00a0\u00a0 BCA\u00a0\u00a0\u00a0\u00a0 CAB\u00a0\u00a0\u00a0\u00a0 CBA<\/p>\n<p>How can we be sure that we have counted them all? We are really just choosing 3 people out of 3, so there are 3 \u00b7 2 \u00b7 1 = 6 ways to do this; we didn&#8217;t really need to list them all. We can just use permutations!<\/p>\n<p>So, out of the 13,800 ways to select 3 people out of 25, six of them involve Abe, Bea and Cindy. The same argument works for any other group of three people (say Abe, Bea and David or Frank, Gloria and Hildy) so each three-person group is counted <em>six times<\/em>. Thus the 13,800 figure is six times too big. The number of distinct three-person groups will be 13,800\/6 = 2300.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>We can generalize the situation in this example above to any problem of choosing a collection of items <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>. If we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by [latex]\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex], and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing <em>r<\/em> items out of <em>n<\/em> possibilities <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em>.<\/p>\n<div class=\"textbox\">\n<h3>Combinations<\/h3>\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[\/latex]<\/p>\n<p>We say that there are <em>nCr<\/em> <strong>combinations<\/strong> of size <em>r<\/em> that may be selected from among <em>n<\/em> choices <em>without replacement<\/em> where <em>order doesn\u2019t matter<\/em>.<\/p>\n<p>We can also write the combinations formula in terms of factorials:<\/p>\n<p style=\"text-align: center;\">[latex]{}_{n}{{C}_{r}}=\\frac{n!}{(n-r)!r!}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q125383\">Show Solution<\/span><\/p>\n<div id=\"q125383\" class=\"hidden-answer\" style=\"display: none\">Since we are choosing 4 people out of 35 <em>without replacement<\/em> where the <em>order of selection is <strong>not<\/strong> important<\/em> there are [latex]{}_{35}{{C}_{4}}=\\frac{35\\cdot34\\cdot33\\cdot32}{4\\cdot3\\cdot2\\cdot1}[\/latex] = 52,360 combinations.<\/div>\n<\/div>\n<p>View the following for more explanation of the combinations examples.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Combinations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/W8kd4YosbzE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?<\/p>\n<\/div>\n<p>In the preceding Try It problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee.\u00a0 (The same of course would be true if the Democrats were in control.) So let&#8217;s consider the problem again, in a slightly more complicated form:<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205320\">Show Solution<\/span><\/p>\n<div id=\"q205320\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this case we need to choose 10 of the 15 Republicans and 9 of the 14 Democrats.\u00a0 There are 15<em>C<\/em>10 = 3003 ways to choose the 10 Republicans and 14<em>C<\/em>9 = 2002 ways to choose the 9 Democrats. But now what?\u00a0 How do we finish the problem?<\/p>\n<p>Suppose we listed all of the possible 10-member Republican groups on 3003 slips of red paper and \u00a0all of the possible 9-member Democratic groups on 2002 slips of blue paper.\u00a0 How many ways can we choose one red slip and one blue slip?\u00a0 This is a job for the Basic Counting Rule!\u00a0 We are simply making one choice from the first category and one choice from the second category, just like in the restaurant menu problems from earlier.<\/p>\n<p>There must be 3003 \u00b7 2002 = 6,012,006 possible ways of selecting the members of the Defense Subcommittee.<\/p>\n<\/div>\n<\/div>\n<p>This example is worked through below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Combinations 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Xqc2sdYN7xo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2466\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: Lippman, David. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Lippman, David\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"2d60a362-3c4b-48e0-adac-3af70abbd2be","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2466","chapter","type-chapter","status-publish","hentry"],"part":329,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2466","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2466\/revisions"}],"predecessor-version":[{"id":3042,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2466\/revisions\/3042"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/parts\/329"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapters\/2466\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/media?parent=2466"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/pressbooks\/v2\/chapter-type?post=2466"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/contributor?post=2466"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/wp-json\/wp\/v2\/license?post=2466"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}