{"id":14826,"date":"2018-09-27T18:37:04","date_gmt":"2018-09-27T18:37:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/counting-principles\/"},"modified":"2019-09-09T21:30:38","modified_gmt":"2019-09-09T21:30:38","slug":"counting-principles","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/counting-principles\/","title":{"raw":"Counting Principles","rendered":"Counting Principles"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Solve counting problems using the Addition Principle.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve counting problems using the Multiplication Principle.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve counting problems using permutations involving n distinct objects.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve counting problems using combinations.<\/li>\r\n \t<li style=\"font-weight: 400;\">Find the number of subsets of a given set.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve counting problems using permutations involving n non-distinct objects.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using the Addition and Multiplication Principles<\/h2>\r\nThe company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The <strong>Addition Principle<\/strong> tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 1.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183634\/CNX_Precalc_Figure_11_05_001n2.jpg\" alt=\"The addition of 3 iPods and 4 iPhones.\" width=\"487\" height=\"358\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Addition Principle<\/h3>\r\nAccording to the <strong>Addition Principle<\/strong>, if one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first <em>or<\/em> second event can occur in [latex]m+n[\/latex] ways.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Using the Addition Principle<\/h3>\r\nThere are 2 vegetarian entr\u00e9e options and 5 meat entr\u00e9e options on a dinner menu. What is the total number of entr\u00e9e options?\r\n\r\n[reveal-answer q=\"507128\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"507128\"]\r\n\r\nWe can add the number of vegetarian options to the number of meat options to find the total number of entr\u00e9e options.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183636\/CNX_Precalc_Figure_11_05_0022.jpg\" alt=\"The addition of the type of options for an entree.\" \/>\r\n\r\nThere are 7 total options.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop computers. What is the total number of computer options?\r\n\r\n[reveal-answer q=\"352795\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"352795\"]\r\n\r\n7\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[ohm_question hide_question_numbers=1]23739[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using the Multiplication Principle<\/h2>\r\nThe <strong>Multiplication Principle<\/strong> applies when we are making more than one selection. Suppose we are choosing an appetizer, an entr\u00e9e, and a dessert. If there are 2 appetizer options, 3 entr\u00e9e options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram in Figure 2.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183638\/CNX_Precalc_Figure_11_05_0032.jpg\" alt=\"A tree diagram of the different menu combinations.\" width=\"975\" height=\"287\" \/>\r\n\r\nThe possible choices are:\r\n<ol>\r\n \t<li>soup, chicken, cake<\/li>\r\n \t<li>soup, chicken, pudding<\/li>\r\n \t<li>soup, fish, cake<\/li>\r\n \t<li>soup, fish, pudding<\/li>\r\n \t<li>soup, steak, cake<\/li>\r\n \t<li>soup, steak, pudding<\/li>\r\n \t<li>salad, chicken, cake<\/li>\r\n \t<li>salad, chicken, pudding<\/li>\r\n \t<li>salad, fish, cake<\/li>\r\n \t<li>salad, fish, pudding<\/li>\r\n \t<li>salad, steak, cake<\/li>\r\n \t<li>salad, steak, pudding<\/li>\r\n<\/ol>\r\nWe can also find the total number of possible dinners by multiplying.\r\n\r\nWe could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle.\r\n<div style=\"text-align: center;\"><a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/10\/Screen-Shot-2015-10-22-at-10.01.45-AM.png\"><img class=\"aligncenter wp-image-12344 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183640\/Screen-Shot-2015-10-22-at-10.01.45-AM.png\" alt=\"Number of appetizer options (2) times number of entree options (3) times number of dessert options (2)\" width=\"663\" height=\"64\" \/><\/a><\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Multiplication Principle<\/h3>\r\nAccording to the <strong>Multiplication Principle<\/strong>, if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways. This is also known as the <strong>Fundamental Counting Principle<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Using the Multiplication Principle<\/h3>\r\nDiane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits.\r\n\r\n[reveal-answer q=\"423662\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"423662\"]\r\n\r\nTo find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183642\/CNX_Precalc_Figure_11_05_0042.jpg\" alt=\"The multiplication of number of skirt options (2) times the number of blouse options (4) times the number of sweater options (2) which equals 16.\" width=\"487\" height=\"82\" \/>\r\n\r\nThere are 16 possible outfits.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials.\r\n\r\n[reveal-answer q=\"520820\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"520820\"]\r\n\r\nThere are 60 possible breakfast specials.\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[ohm_question hide_question_numbers=1]172904[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding the Number of Permutations of n Distinct Objects<\/h2>\r\nThe Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a <strong>permutation<\/strong>.\r\n<h2>Finding the Number of Permutations of <em>n<\/em> Distinct Objects Using the Multiplication Principle<\/h2>\r\nTo solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183644\/CNX_Precalc_Figure_11_05_0052.jpg\" alt=\"\" \/>\r\n\r\nThere are four options for the first place, so we write a 4 on the first line.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183646\/CNX_Precalc_Figure_11_05_0062.jpg\" alt=\"Four times two blanks spots.\" \/>\r\n\r\nAfter the first place has been filled, there are three options for the second place so we write a 3 on the second line.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183648\/CNX_Precalc_Figure_11_05_0072.jpg\" alt=\"Four times three times one blank spot.\" \/>\r\n\r\nAfter the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183650\/CNX_Precalc_Figure_11_05_0082.jpg\" alt=\"\" \/>\r\n\r\nThere are 24 possible permutations of the paintings.\r\n<div class=\"textbox\">\r\n<h3>How To: Given [latex]n[\/latex] distinct options, determine how many permutations there are.<\/h3>\r\n<ol>\r\n \t<li>Determine how many options there are for the first situation.<\/li>\r\n \t<li>Determine how many options are left for the second situation.<\/li>\r\n \t<li>Continue until all of the spots are filled.<\/li>\r\n \t<li>Multiply the numbers together.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding the Number of Permutations Using the Multiplication Principle<\/h3>\r\nAt a swimming competition, nine swimmers compete in a race.\r\n<ol>\r\n \t<li>How many ways can they place first, second, and third?<\/li>\r\n \t<li>How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.)<\/li>\r\n \t<li>How many ways can all nine swimmers line up for a photo?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"203736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203736\"]\r\n<ol>\r\n \t<li>Draw lines for each place.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183652\/CNX_Precalc_Figure_11_05_0092.jpg\" alt=\"\" \/>There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183654\/CNX_Precalc_Figure_11_05_0102.jpg\" alt=\"\" \/>Multiply to find that there are 504 ways for the swimmers to place.<\/li>\r\n \t<li>Draw lines for describing each place.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183656\/CNX_Precalc_Figure_11_05_0112.jpg\" alt=\"\" \/>We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183658\/CNX_Precalc_Figure_11_05_0122.jpg\" alt=\"\" \/>Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first.<\/li>\r\n \t<li>Draw lines for describing each place in the photo.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183700\/CNX_Precalc_Figure_11_05_0132.jpg\" alt=\"\" \/>There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot.<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183703\/CNX_Precalc_Figure_11_05_0142.jpg\" alt=\"\" \/>There are 362,880 possible permutations for the swimmers to line up.<\/li>\r\n<\/ol>\r\n<div>\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of [latex]n[\/latex] distinct objects can always be found by [latex]n![\/latex].\r\n\r\n<\/div>\r\nA family of five is having portraits taken. Use the Multiplication Principle to find the following.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It \\4<\/h3>\r\nA family has 2 parents and 3 children. How many ways can the family line up for the portrait?\r\n\r\n[reveal-answer q=\"599266\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"599266\"]120[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nHow many ways can the photographer line up 3 of the family members?\r\n\r\n[reveal-answer q=\"270763\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"270763\"]60[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nHow many ways can the family line up for the portrait if the parents stand on each end?\r\n\r\n[reveal-answer q=\"999671\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"999671\"]12[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172908[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Finding the Number of Permutations of <em>n<\/em> Distinct Objects Using a Formula<\/h2>\r\nFor some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let\u2019s look at two common notations for permutations. If we have a set of [latex]n[\/latex] objects and we want to choose [latex]r[\/latex] objects from the set in order, we write [latex]P\\left(n,r\\right)[\/latex]. Another way to write this is [latex]{n}_{}{P}_{r}[\/latex], a notation commonly seen on computers and calculators. To calculate [latex]P\\left(n,r\\right)[\/latex], we begin by finding [latex]n![\/latex], the number of ways to line up all [latex]n[\/latex] objects. We then divide by [latex]\\left(n-r\\right)![\/latex] to cancel out the [latex]\\left(n-r\\right)[\/latex] items that we do not wish to line up.\r\n\r\nLet\u2019s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is [latex]6\\times 5\\times 4=120[\/latex]. Using factorials, we get the same result.\r\n<div style=\"text-align: center;\">[latex]\\frac{6!}{3!}=\\frac{6\\cdot 5\\cdot 4\\cdot 3!}{3!}=6\\cdot 5\\cdot 4=120[\/latex]<\/div>\r\nThere are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.\r\n<div style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/div>\r\nNote that the formula stills works if we are choosing <u>all<\/u> [latex]n[\/latex] objects and placing them in order. In that case we would be dividing by [latex]\\left(n-n\\right)![\/latex] or [latex]0![\/latex], which we said earlier is equal to 1. So the number of permutations of [latex]n[\/latex] objects taken [latex]n[\/latex] at a time is [latex]\\frac{n!}{1}[\/latex] or just [latex]n!\\text{.}[\/latex]\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for Permutations of <em>n<\/em> Distinct Objects<\/h3>\r\nGiven [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set in order is\r\n<div style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a word problem, evaluate the possible permutations.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]n[\/latex] from the given information.<\/li>\r\n \t<li>Identify [latex]r[\/latex] from the given information.<\/li>\r\n \t<li>Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\r\n \t<li>Evaluate.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Finding the Number of Permutations Using the Formula<\/h3>\r\nA professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?\r\n\r\n[reveal-answer q=\"977518\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977518\"]\r\n\r\nSubstitute [latex]n=12[\/latex] and [latex]r=9[\/latex] into the permutation formula and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(n,r\\right)&amp;=\\frac{n!}{\\left(n-r\\right)!}\\\\ P\\left(12,9\\right)&amp;=\\frac{12!}{\\left(12 - 9\\right)!}=\\frac{12!}{3!}=79\\text{,}833\\text{,}600 \\end{align}[\/latex]<\/p>\r\nThere are 79,833,600 possible permutations of exam questions.\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can also use a calculator to find permutations. For this problem, we would enter 15, press the [latex]{}_{n}{P}_{r}[\/latex]\u00a0function, enter 12, and then press the equal sign. The [latex]{}_{n}{P}_{r}[\/latex]\u00a0function may be located under the MATH menu with probability commands.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Could we have solved \u00a0using the Multiplication Principle?<\/h3>\r\n<em>Yes. We could have multiplied<\/em> [latex]15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6\\cdot 5\\cdot 4[\/latex] <em>to find the same answer<\/em>.\r\n\r\n<\/div>\r\nA play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nHow many ways can the 7 actors line up?\r\n\r\n[reveal-answer q=\"863346\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"863346\"]\r\n\r\n[latex]P\\left(7,7\\right)=5,040[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nHow many ways can 5 of the 7 actors be chosen to line up?\r\n\r\n[reveal-answer q=\"994559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"994559\"]\r\n\r\n[latex]P\\left(7,5\\right)=2,520[\/latex]\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[ohm_question hide_question_numbers=1]170517[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Find the Number of Combinations Using the Formula<\/h2>\r\nSo far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with <strong>combinations<\/strong>. A selection of [latex]r[\/latex] objects from a set of [latex]n[\/latex] objects where the order does not matter can be written as [latex]C\\left(n,r\\right)[\/latex]. Just as with permutations, [latex]\\text{C}\\left(n,r\\right)[\/latex] can also be written as [latex]{}_{n}{C}_{r}[\/latex]. In this case, the general formula is as follows.\r\n<div style=\"text-align: center;\">[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\r\nAn earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order 3 paintings. There are [latex]\\frac{24}{6}[\/latex], or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are <em>not<\/em> selecting 1 painting. There are 4 paintings we could choose <em>not<\/em> to select, so there are 4 ways to select 3 of the 4 paintings.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for Combinations of <em>n<\/em> Distinct Objects<\/h3>\r\nGiven [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is\r\n<div style=\"text-align: center;\">[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a number of options, determine the possible number of combinations.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]n[\/latex] from the given information.<\/li>\r\n \t<li>Identify [latex]r[\/latex] from the given information.<\/li>\r\n \t<li>Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\r\n \t<li>Evaluate.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Finding the Number of Combinations Using the Formula<\/h3>\r\nA fast food restaurant offers five side dish options. Your meal comes with two side dishes.\r\n<ol>\r\n \t<li>How many ways can you select your side dishes?<\/li>\r\n \t<li>How many ways can you select 3 side dishes?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"538399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538399\"]\r\n<ol>\r\n \t<li>We want to choose 2 side dishes from 5 options.\r\n<div>[latex]\\text{C}\\left(5,2\\right)=\\frac{5!}{2!\\left(5 - 2\\right)!}=10[\/latex]<\/div><\/li>\r\n \t<li>We want to choose 3 side dishes from 5 options.\r\n<div>[latex]\\text{C}\\left(5,3\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=10[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can also use a graphing calculator to find combinations. Enter 5, then press [latex]{}_{n}{C}_{r}[\/latex], enter 3, and then press the equal sign. The [latex]{}_{n}{C}_{r}[\/latex], function may be located under the MATH menu with probability commands.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Is it a coincidence that parts (a) and (b) in Example 4\u00a0have the same answers?<\/h3>\r\n<em>No. When we choose r objects from n objects, we are <strong>not<\/strong> choosing [latex]\\left(n-r\\right)[\/latex] objects. Therefore, [latex]C\\left(n,r\\right)=C\\left(n,n-r\\right)[\/latex]. <\/em>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nAn ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?\r\n\r\n[reveal-answer q=\"759484\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759484\"]\r\n\r\n[latex]C\\left(10,3\\right)=120[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Number of Subsets of a Set<\/h2>\r\nWe have looked only at combination problems in which we chose exactly [latex]r[\/latex] objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible?\r\n\r\nTo answer this question, we need to consider pizzas with any number of toppings. There is [latex]C\\left(5,0\\right)=1[\/latex] way to order a pizza with no toppings. There are [latex]C\\left(5,1\\right)=5[\/latex] ways to order a pizza with exactly one topping. If we continue this process, we get\r\n<div style=\"text-align: center;\">[latex]C\\left(5,0\\right)+C\\left(5,1\\right)+C\\left(5,2\\right)+C\\left(5,3\\right)+C\\left(5,4\\right)+C\\left(5,5\\right)=32[\/latex]<\/div>\r\nThere are 32 possible pizzas. This result is equal to [latex]{2}^{5}[\/latex].\r\n\r\nWe are presented with a sequence of choices. For each of the [latex]n[\/latex] objects we have two choices: include it in the subset or not. So for the whole subset we have made [latex]n[\/latex] choices, each with two options. So there are a total of [latex]2\\cdot 2\\cdot 2\\cdot \\dots \\cdot 2[\/latex] possible resulting subsets, all the way from the empty subset, which we obtain when we say \"no\" each time, to the original set itself, which we obtain when we say \"yes\" each time.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Number of Subsets of a Set<\/h3>\r\nA set containing <em>n<\/em> distinct objects has [latex]{2}^{n}[\/latex] subsets.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Finding the Number of Subsets of a Set<\/h3>\r\nA restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?\r\n\r\n[reveal-answer q=\"961472\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"961472\"]\r\n\r\nWe are looking for the number of subsets of a set with 4 objects. Substitute [latex]n=4[\/latex] into the formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{2}^{n}&amp;={2}^{4} \\\\ &amp;=16 \\end{align}[\/latex]<\/p>\r\nThere are 16 possible ways to order a potato.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?\r\n\r\n[reveal-answer q=\"710378\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"710378\"]\r\n\r\n64 sundaes\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding the Number of Permutations of n Non-Distinct Objects<\/h2>\r\nWe have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be [latex]12![\/latex] ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the [latex]12![\/latex] permutations we counted are duplicates. The general formula for this situation is as follows.\r\n<div style=\"text-align: center;\">[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/div>\r\nIn this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are [latex]4![\/latex] ways to order the stars and [latex]3![\/latex] ways to order the moon.\r\n<div style=\"text-align: center;\">[latex]\\frac{12!}{4!3!}=3\\text{,}326\\text{,}400[\/latex]<\/div>\r\nThere are 3,326,400 ways to order the sheet of stickers.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for Finding the Number of Permutations of <em>n<\/em> Non-Distinct Objects<\/h3>\r\nIf there are [latex]n[\/latex] elements in a set and [latex]{r}_{1}[\/latex] are alike, [latex]{r}_{2}[\/latex] are alike, [latex]{r}_{3}[\/latex] are alike, and so on through [latex]{r}_{k}[\/latex], the number of permutations can be found by\r\n<p style=\"text-align: center;\">[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Finding the Number of Permutations of <em>n<\/em> Non-Distinct Objects<\/h3>\r\nFind the number of rearrangements of the letters in the word DISTINCT.\r\n\r\n[reveal-answer q=\"475893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"475893\"]\r\n\r\nThere are 8 letters. Both I and T are repeated 2 times. Substitute [latex]n=8, {r}_{1}=2, [\/latex] and [latex] {r}_{2}=2 [\/latex] into the formula.\r\n<p style=\"text-align: center;\">[latex]\\frac{8!}{2!2!}=10\\text{,}080 [\/latex]<\/p>\r\nThere are 10,080 arrangements.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind the number of rearrangements of the letters in the word CARRIER.\r\n\r\n[reveal-answer q=\"587460\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"587460\"]840[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]107963[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>number of permutations of [latex]n[\/latex] distinct objects taken [latex]r[\/latex] at a time<\/td>\r\n<td>[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>number of combinations of [latex]n[\/latex] distinct objects taken [latex]r[\/latex] at a time<\/td>\r\n<td>[latex]C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>number of permutations of [latex]n[\/latex] non-distinct objects<\/td>\r\n<td>[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>If one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first or second event can occur in [latex]m+n[\/latex] ways.<\/li>\r\n \t<li>If one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways.<\/li>\r\n \t<li>A permutation is an ordering of [latex]n[\/latex] objects.<\/li>\r\n \t<li>If we have a set of [latex]n[\/latex] objects and we want to choose [latex]r[\/latex] objects from the set in order, we write [latex]P\\left(n,r\\right)[\/latex].<\/li>\r\n \t<li>Permutation problems can be solved using the Multiplication Principle or the formula for [latex]P\\left(n,r\\right)[\/latex].<\/li>\r\n \t<li>A selection of objects where the order does not matter is a combination.<\/li>\r\n \t<li>Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is [latex]\\text{C}\\left(n,r\\right)[\/latex] and can be found using a formula.<\/li>\r\n \t<li>A set containing [latex]n[\/latex] distinct objects has [latex]{2}^{n}[\/latex] subsets.<\/li>\r\n \t<li>For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135255272\" class=\"definition\">\r\n \t<dt>Addition Principle<\/dt>\r\n \t<dd id=\"fs-id1165135255277\">if one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first or second event can occur in [latex]m+n[\/latex] ways<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137645178\" class=\"definition\">\r\n \t<dt>combination<\/dt>\r\n \t<dd id=\"fs-id1165135160429\">a selection of objects in which order does not matter<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135160433\" class=\"definition\">\r\n \t<dt>Fundamental Counting Principle<\/dt>\r\n \t<dd id=\"fs-id1165137668328\">if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways; also known as the Multiplication Principle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137651693\" class=\"definition\">\r\n \t<dt>Multiplication Principle<\/dt>\r\n \t<dd id=\"fs-id1165137651698\">if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways; also known as the Fundamental Counting Principle<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137871530\" class=\"definition\">\r\n \t<dt>permutation<\/dt>\r\n \t<dd id=\"fs-id1165137551518\">a selection of objects in which order matters<\/dd>\r\n<\/dl>\r\n&nbsp;\r\n\r\nFor the following exercises, assume that there are [latex]n[\/latex] ways an event [latex]A[\/latex] can happen, [latex]m[\/latex] ways an event [latex]B[\/latex] can happen, and that [latex]A\\text{ and }B[\/latex] are non-overlapping.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li style=\"font-weight: 400;\">Solve counting problems using the Addition Principle.<\/li>\n<li style=\"font-weight: 400;\">Solve counting problems using the Multiplication Principle.<\/li>\n<li style=\"font-weight: 400;\">Solve counting problems using permutations involving n distinct objects.<\/li>\n<li style=\"font-weight: 400;\">Solve counting problems using combinations.<\/li>\n<li style=\"font-weight: 400;\">Find the number of subsets of a given set.<\/li>\n<li style=\"font-weight: 400;\">Solve counting problems using permutations involving n non-distinct objects.<\/li>\n<\/ul>\n<\/div>\n<h2>Using the Addition and Multiplication Principles<\/h2>\n<p>The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The <strong>Addition Principle<\/strong> tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 1.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183634\/CNX_Precalc_Figure_11_05_001n2.jpg\" alt=\"The addition of 3 iPods and 4 iPhones.\" width=\"487\" height=\"358\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Addition Principle<\/h3>\n<p>According to the <strong>Addition Principle<\/strong>, if one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first <em>or<\/em> second event can occur in [latex]m+n[\/latex] ways.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Using the Addition Principle<\/h3>\n<p>There are 2 vegetarian entr\u00e9e options and 5 meat entr\u00e9e options on a dinner menu. What is the total number of entr\u00e9e options?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q507128\">Show Solution<\/span><\/p>\n<div id=\"q507128\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can add the number of vegetarian options to the number of meat options to find the total number of entr\u00e9e options.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183636\/CNX_Precalc_Figure_11_05_0022.jpg\" alt=\"The addition of the type of options for an entree.\" \/><\/p>\n<p>There are 7 total options.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop computers. What is the total number of computer options?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q352795\">Show Solution<\/span><\/p>\n<div id=\"q352795\" class=\"hidden-answer\" style=\"display: none\">\n<p>7<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm23739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23739&theme=oea&iframe_resize_id=ohm23739\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Multiplication Principle<\/h2>\n<p>The <strong>Multiplication Principle<\/strong> applies when we are making more than one selection. Suppose we are choosing an appetizer, an entr\u00e9e, and a dessert. If there are 2 appetizer options, 3 entr\u00e9e options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram in Figure 2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183638\/CNX_Precalc_Figure_11_05_0032.jpg\" alt=\"A tree diagram of the different menu combinations.\" width=\"975\" height=\"287\" \/><\/p>\n<p>The possible choices are:<\/p>\n<ol>\n<li>soup, chicken, cake<\/li>\n<li>soup, chicken, pudding<\/li>\n<li>soup, fish, cake<\/li>\n<li>soup, fish, pudding<\/li>\n<li>soup, steak, cake<\/li>\n<li>soup, steak, pudding<\/li>\n<li>salad, chicken, cake<\/li>\n<li>salad, chicken, pudding<\/li>\n<li>salad, fish, cake<\/li>\n<li>salad, fish, pudding<\/li>\n<li>salad, steak, cake<\/li>\n<li>salad, steak, pudding<\/li>\n<\/ol>\n<p>We can also find the total number of possible dinners by multiplying.<\/p>\n<p>We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle.<\/p>\n<div style=\"text-align: center;\"><a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/10\/Screen-Shot-2015-10-22-at-10.01.45-AM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-12344 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183640\/Screen-Shot-2015-10-22-at-10.01.45-AM.png\" alt=\"Number of appetizer options (2) times number of entree options (3) times number of dessert options (2)\" width=\"663\" height=\"64\" \/><\/a><\/div>\n<div class=\"textbox\">\n<h3>A General Note: The Multiplication Principle<\/h3>\n<p>According to the <strong>Multiplication Principle<\/strong>, if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways. This is also known as the <strong>Fundamental Counting Principle<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Using the Multiplication Principle<\/h3>\n<p>Diane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q423662\">Show Solution<\/span><\/p>\n<div id=\"q423662\" class=\"hidden-answer\" style=\"display: none\">\n<p>To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183642\/CNX_Precalc_Figure_11_05_0042.jpg\" alt=\"The multiplication of number of skirt options (2) times the number of blouse options (4) times the number of sweater options (2) which equals 16.\" width=\"487\" height=\"82\" \/><\/p>\n<p>There are 16 possible outfits.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q520820\">Show Solution<\/span><\/p>\n<div id=\"q520820\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 60 possible breakfast specials.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172904\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172904&theme=oea&iframe_resize_id=ohm172904\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding the Number of Permutations of n Distinct Objects<\/h2>\n<p>The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a <strong>permutation<\/strong>.<\/p>\n<h2>Finding the Number of Permutations of <em>n<\/em> Distinct Objects Using the Multiplication Principle<\/h2>\n<p>To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183644\/CNX_Precalc_Figure_11_05_0052.jpg\" alt=\"\" \/><\/p>\n<p>There are four options for the first place, so we write a 4 on the first line.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183646\/CNX_Precalc_Figure_11_05_0062.jpg\" alt=\"Four times two blanks spots.\" \/><\/p>\n<p>After the first place has been filled, there are three options for the second place so we write a 3 on the second line.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183648\/CNX_Precalc_Figure_11_05_0072.jpg\" alt=\"Four times three times one blank spot.\" \/><\/p>\n<p>After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183650\/CNX_Precalc_Figure_11_05_0082.jpg\" alt=\"\" \/><\/p>\n<p>There are 24 possible permutations of the paintings.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given [latex]n[\/latex] distinct options, determine how many permutations there are.<\/h3>\n<ol>\n<li>Determine how many options there are for the first situation.<\/li>\n<li>Determine how many options are left for the second situation.<\/li>\n<li>Continue until all of the spots are filled.<\/li>\n<li>Multiply the numbers together.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding the Number of Permutations Using the Multiplication Principle<\/h3>\n<p>At a swimming competition, nine swimmers compete in a race.<\/p>\n<ol>\n<li>How many ways can they place first, second, and third?<\/li>\n<li>How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.)<\/li>\n<li>How many ways can all nine swimmers line up for a photo?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203736\">Show Solution<\/span><\/p>\n<div id=\"q203736\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Draw lines for each place.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183652\/CNX_Precalc_Figure_11_05_0092.jpg\" alt=\"\" \/>There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183654\/CNX_Precalc_Figure_11_05_0102.jpg\" alt=\"\" \/>Multiply to find that there are 504 ways for the swimmers to place.<\/li>\n<li>Draw lines for describing each place.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183656\/CNX_Precalc_Figure_11_05_0112.jpg\" alt=\"\" \/>We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183658\/CNX_Precalc_Figure_11_05_0122.jpg\" alt=\"\" \/>Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first.<\/li>\n<li>Draw lines for describing each place in the photo.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183700\/CNX_Precalc_Figure_11_05_0132.jpg\" alt=\"\" \/>There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot.<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183703\/CNX_Precalc_Figure_11_05_0142.jpg\" alt=\"\" \/>There are 362,880 possible permutations for the swimmers to line up.<\/li>\n<\/ol>\n<div>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of [latex]n[\/latex] distinct objects can always be found by [latex]n![\/latex].<\/p>\n<\/div>\n<p>A family of five is having portraits taken. Use the Multiplication Principle to find the following.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It \\4<\/h3>\n<p>A family has 2 parents and 3 children. How many ways can the family line up for the portrait?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q599266\">Show Solution<\/span><\/p>\n<div id=\"q599266\" class=\"hidden-answer\" style=\"display: none\">120<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>How many ways can the photographer line up 3 of the family members?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q270763\">Show Solution<\/span><\/p>\n<div id=\"q270763\" class=\"hidden-answer\" style=\"display: none\">60<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>How many ways can the family line up for the portrait if the parents stand on each end?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q999671\">Show Solution<\/span><\/p>\n<div id=\"q999671\" class=\"hidden-answer\" style=\"display: none\">12<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172908\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172908&theme=oea&iframe_resize_id=ohm172908\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Finding the Number of Permutations of <em>n<\/em> Distinct Objects Using a Formula<\/h2>\n<p>For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let\u2019s look at two common notations for permutations. If we have a set of [latex]n[\/latex] objects and we want to choose [latex]r[\/latex] objects from the set in order, we write [latex]P\\left(n,r\\right)[\/latex]. Another way to write this is [latex]{n}_{}{P}_{r}[\/latex], a notation commonly seen on computers and calculators. To calculate [latex]P\\left(n,r\\right)[\/latex], we begin by finding [latex]n![\/latex], the number of ways to line up all [latex]n[\/latex] objects. We then divide by [latex]\\left(n-r\\right)![\/latex] to cancel out the [latex]\\left(n-r\\right)[\/latex] items that we do not wish to line up.<\/p>\n<p>Let\u2019s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is [latex]6\\times 5\\times 4=120[\/latex]. Using factorials, we get the same result.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{6!}{3!}=\\frac{6\\cdot 5\\cdot 4\\cdot 3!}{3!}=6\\cdot 5\\cdot 4=120[\/latex]<\/div>\n<p>There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.<\/p>\n<div style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/div>\n<p>Note that the formula stills works if we are choosing <u>all<\/u> [latex]n[\/latex] objects and placing them in order. In that case we would be dividing by [latex]\\left(n-n\\right)![\/latex] or [latex]0![\/latex], which we said earlier is equal to 1. So the number of permutations of [latex]n[\/latex] objects taken [latex]n[\/latex] at a time is [latex]\\frac{n!}{1}[\/latex] or just [latex]n!\\text{.}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for Permutations of <em>n<\/em> Distinct Objects<\/h3>\n<p>Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set in order is<\/p>\n<div style=\"text-align: center;\">[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a word problem, evaluate the possible permutations.<\/h3>\n<ol>\n<li>Identify [latex]n[\/latex] from the given information.<\/li>\n<li>Identify [latex]r[\/latex] from the given information.<\/li>\n<li>Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\n<li>Evaluate.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Finding the Number of Permutations Using the Formula<\/h3>\n<p>A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q977518\">Show Solution<\/span><\/p>\n<div id=\"q977518\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]n=12[\/latex] and [latex]r=9[\/latex] into the permutation formula and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}P\\left(n,r\\right)&=\\frac{n!}{\\left(n-r\\right)!}\\\\ P\\left(12,9\\right)&=\\frac{12!}{\\left(12 - 9\\right)!}=\\frac{12!}{3!}=79\\text{,}833\\text{,}600 \\end{align}[\/latex]<\/p>\n<p>There are 79,833,600 possible permutations of exam questions.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can also use a calculator to find permutations. For this problem, we would enter 15, press the [latex]{}_{n}{P}_{r}[\/latex]\u00a0function, enter 12, and then press the equal sign. The [latex]{}_{n}{P}_{r}[\/latex]\u00a0function may be located under the MATH menu with probability commands.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Could we have solved \u00a0using the Multiplication Principle?<\/h3>\n<p><em>Yes. We could have multiplied<\/em> [latex]15\\cdot 14\\cdot 13\\cdot 12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6\\cdot 5\\cdot 4[\/latex] <em>to find the same answer<\/em>.<\/p>\n<\/div>\n<p>A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>How many ways can the 7 actors line up?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q863346\">Show Solution<\/span><\/p>\n<div id=\"q863346\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]P\\left(7,7\\right)=5,040[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>How many ways can 5 of the 7 actors be chosen to line up?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q994559\">Show Solution<\/span><\/p>\n<div id=\"q994559\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]P\\left(7,5\\right)=2,520[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm170517\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=170517&theme=oea&iframe_resize_id=ohm170517\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Find the Number of Combinations Using the Formula<\/h2>\n<p>So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with <strong>combinations<\/strong>. A selection of [latex]r[\/latex] objects from a set of [latex]n[\/latex] objects where the order does not matter can be written as [latex]C\\left(n,r\\right)[\/latex]. Just as with permutations, [latex]\\text{C}\\left(n,r\\right)[\/latex] can also be written as [latex]{}_{n}{C}_{r}[\/latex]. In this case, the general formula is as follows.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n<p>An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are [latex]3!=3\\cdot 2\\cdot 1=6[\/latex] ways to order 3 paintings. There are [latex]\\frac{24}{6}[\/latex], or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are <em>not<\/em> selecting 1 painting. There are 4 paintings we could choose <em>not<\/em> to select, so there are 4 ways to select 3 of the 4 paintings.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for Combinations of <em>n<\/em> Distinct Objects<\/h3>\n<p>Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is<\/p>\n<div style=\"text-align: center;\">[latex]\\text{C}\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a number of options, determine the possible number of combinations.<\/h3>\n<ol>\n<li>Identify [latex]n[\/latex] from the given information.<\/li>\n<li>Identify [latex]r[\/latex] from the given information.<\/li>\n<li>Replace [latex]n[\/latex] and [latex]r[\/latex] in the formula with the given values.<\/li>\n<li>Evaluate.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Finding the Number of Combinations Using the Formula<\/h3>\n<p>A fast food restaurant offers five side dish options. Your meal comes with two side dishes.<\/p>\n<ol>\n<li>How many ways can you select your side dishes?<\/li>\n<li>How many ways can you select 3 side dishes?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538399\">Show Solution<\/span><\/p>\n<div id=\"q538399\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>We want to choose 2 side dishes from 5 options.\n<div>[latex]\\text{C}\\left(5,2\\right)=\\frac{5!}{2!\\left(5 - 2\\right)!}=10[\/latex]<\/div>\n<\/li>\n<li>We want to choose 3 side dishes from 5 options.\n<div>[latex]\\text{C}\\left(5,3\\right)=\\frac{5!}{3!\\left(5 - 3\\right)!}=10[\/latex]<\/div>\n<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p>We can also use a graphing calculator to find combinations. Enter 5, then press [latex]{}_{n}{C}_{r}[\/latex], enter 3, and then press the equal sign. The [latex]{}_{n}{C}_{r}[\/latex], function may be located under the MATH menu with probability commands.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Is it a coincidence that parts (a) and (b) in Example 4\u00a0have the same answers?<\/h3>\n<p><em>No. When we choose r objects from n objects, we are <strong>not<\/strong> choosing [latex]\\left(n-r\\right)[\/latex] objects. Therefore, [latex]C\\left(n,r\\right)=C\\left(n,n-r\\right)[\/latex]. <\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q759484\">Show Solution<\/span><\/p>\n<div id=\"q759484\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]C\\left(10,3\\right)=120[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Number of Subsets of a Set<\/h2>\n<p>We have looked only at combination problems in which we chose exactly [latex]r[\/latex] objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible?<\/p>\n<p>To answer this question, we need to consider pizzas with any number of toppings. There is [latex]C\\left(5,0\\right)=1[\/latex] way to order a pizza with no toppings. There are [latex]C\\left(5,1\\right)=5[\/latex] ways to order a pizza with exactly one topping. If we continue this process, we get<\/p>\n<div style=\"text-align: center;\">[latex]C\\left(5,0\\right)+C\\left(5,1\\right)+C\\left(5,2\\right)+C\\left(5,3\\right)+C\\left(5,4\\right)+C\\left(5,5\\right)=32[\/latex]<\/div>\n<p>There are 32 possible pizzas. This result is equal to [latex]{2}^{5}[\/latex].<\/p>\n<p>We are presented with a sequence of choices. For each of the [latex]n[\/latex] objects we have two choices: include it in the subset or not. So for the whole subset we have made [latex]n[\/latex] choices, each with two options. So there are a total of [latex]2\\cdot 2\\cdot 2\\cdot \\dots \\cdot 2[\/latex] possible resulting subsets, all the way from the empty subset, which we obtain when we say &#8220;no&#8221; each time, to the original set itself, which we obtain when we say &#8220;yes&#8221; each time.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Number of Subsets of a Set<\/h3>\n<p>A set containing <em>n<\/em> distinct objects has [latex]{2}^{n}[\/latex] subsets.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Finding the Number of Subsets of a Set<\/h3>\n<p>A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q961472\">Show Solution<\/span><\/p>\n<div id=\"q961472\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are looking for the number of subsets of a set with 4 objects. Substitute [latex]n=4[\/latex] into the formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{2}^{n}&={2}^{4} \\\\ &=16 \\end{align}[\/latex]<\/p>\n<p>There are 16 possible ways to order a potato.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q710378\">Show Solution<\/span><\/p>\n<div id=\"q710378\" class=\"hidden-answer\" style=\"display: none\">\n<p>64 sundaes<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding the Number of Permutations of n Non-Distinct Objects<\/h2>\n<p>We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be [latex]12![\/latex] ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the [latex]12![\/latex] permutations we counted are duplicates. The general formula for this situation is as follows.<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/div>\n<p>In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are [latex]4![\/latex] ways to order the stars and [latex]3![\/latex] ways to order the moon.<\/p>\n<div style=\"text-align: center;\">[latex]\\frac{12!}{4!3!}=3\\text{,}326\\text{,}400[\/latex]<\/div>\n<p>There are 3,326,400 ways to order the sheet of stickers.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for Finding the Number of Permutations of <em>n<\/em> Non-Distinct Objects<\/h3>\n<p>If there are [latex]n[\/latex] elements in a set and [latex]{r}_{1}[\/latex] are alike, [latex]{r}_{2}[\/latex] are alike, [latex]{r}_{3}[\/latex] are alike, and so on through [latex]{r}_{k}[\/latex], the number of permutations can be found by<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Finding the Number of Permutations of <em>n<\/em> Non-Distinct Objects<\/h3>\n<p>Find the number of rearrangements of the letters in the word DISTINCT.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q475893\">Show Solution<\/span><\/p>\n<div id=\"q475893\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are 8 letters. Both I and T are repeated 2 times. Substitute [latex]n=8, {r}_{1}=2,[\/latex] and [latex]{r}_{2}=2[\/latex] into the formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{8!}{2!2!}=10\\text{,}080[\/latex]<\/p>\n<p>There are 10,080 arrangements.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find the number of rearrangements of the letters in the word CARRIER.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q587460\">Show Solution<\/span><\/p>\n<div id=\"q587460\" class=\"hidden-answer\" style=\"display: none\">840<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm107963\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=107963&theme=oea&iframe_resize_id=ohm107963\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td>number of permutations of [latex]n[\/latex] distinct objects taken [latex]r[\/latex] at a time<\/td>\n<td>[latex]P\\left(n,r\\right)=\\frac{n!}{\\left(n-r\\right)!}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>number of combinations of [latex]n[\/latex] distinct objects taken [latex]r[\/latex] at a time<\/td>\n<td>[latex]C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>number of permutations of [latex]n[\/latex] non-distinct objects<\/td>\n<td>[latex]\\dfrac{n!}{{r}_{1}!{r}_{2}!\\dots {r}_{k}!}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>If one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first or second event can occur in [latex]m+n[\/latex] ways.<\/li>\n<li>If one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways.<\/li>\n<li>A permutation is an ordering of [latex]n[\/latex] objects.<\/li>\n<li>If we have a set of [latex]n[\/latex] objects and we want to choose [latex]r[\/latex] objects from the set in order, we write [latex]P\\left(n,r\\right)[\/latex].<\/li>\n<li>Permutation problems can be solved using the Multiplication Principle or the formula for [latex]P\\left(n,r\\right)[\/latex].<\/li>\n<li>A selection of objects where the order does not matter is a combination.<\/li>\n<li>Given [latex]n[\/latex] distinct objects, the number of ways to select [latex]r[\/latex] objects from the set is [latex]\\text{C}\\left(n,r\\right)[\/latex] and can be found using a formula.<\/li>\n<li>A set containing [latex]n[\/latex] distinct objects has [latex]{2}^{n}[\/latex] subsets.<\/li>\n<li>For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135255272\" class=\"definition\">\n<dt>Addition Principle<\/dt>\n<dd id=\"fs-id1165135255277\">if one event can occur in [latex]m[\/latex] ways and a second event with no common outcomes can occur in [latex]n[\/latex] ways, then the first or second event can occur in [latex]m+n[\/latex] ways<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137645178\" class=\"definition\">\n<dt>combination<\/dt>\n<dd id=\"fs-id1165135160429\">a selection of objects in which order does not matter<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135160433\" class=\"definition\">\n<dt>Fundamental Counting Principle<\/dt>\n<dd id=\"fs-id1165137668328\">if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways; also known as the Multiplication Principle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137651693\" class=\"definition\">\n<dt>Multiplication Principle<\/dt>\n<dd id=\"fs-id1165137651698\">if one event can occur in [latex]m[\/latex] ways and a second event can occur in [latex]n[\/latex] ways after the first event has occurred, then the two events can occur in [latex]m\\times n[\/latex] ways; also known as the Fundamental Counting Principle<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137871530\" class=\"definition\">\n<dt>permutation<\/dt>\n<dd id=\"fs-id1165137551518\">a selection of objects in which order matters<\/dd>\n<\/dl>\n<p>&nbsp;<\/p>\n<p>For the following exercises, assume that there are [latex]n[\/latex] ways an event [latex]A[\/latex] can happen, [latex]m[\/latex] ways an event [latex]B[\/latex] can happen, and that [latex]A\\text{ and }B[\/latex] are non-overlapping.<\/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-14826\">\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>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14826","chapter","type-chapter","status-publish","hentry"],"part":14758,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14826","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14826\/revisions"}],"predecessor-version":[{"id":15859,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14826\/revisions\/15859"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14758"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14826\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=14826"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14826"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=14826"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=14826"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}