{"id":457,"date":"2016-10-12T22:24:26","date_gmt":"2016-10-12T22:24:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=457"},"modified":"2022-12-01T18:29:09","modified_gmt":"2022-12-01T18:29:09","slug":"cardinality","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/mathforliberalartscorequisite\/chapter\/cardinality\/","title":{"raw":"Cardinality","rendered":"Cardinality"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve real-life problems involving sets, subsets, and cardinality properties<\/li>\r\n<\/ul>\r\n<\/div>\r\nOften times we are interested in the number of items in a set or subset. This is called the cardinality of the set.\r\n<div class=\"textbox examples\">\r\n<h3>about the notation<\/h3>\r\nIn the definition of\u00a0<em>cardinality<\/em> below, note that the symbol [latex]{\\lvert}A{\\rvert}[\/latex] looks like absolute value of [latex]A[\/latex] but does not denote absolute value. This symbol would be understood to represent the cardinality of set [latex]A[\/latex] rather than absolute value by the context in which it is used. Note that the symbol n[latex]\\left(A\\right)[\/latex] is also used to represent the cardinality of set [latex]A[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Cardinality<\/h3>\r\nThe number of elements in a set is the cardinality of that set.\r\n\r\nThe cardinality of the set <em>A<\/em> is often notated as [latex]{\\lvert}A{\\rvert}[\/latex]\u00a0or\u00a0n[latex]\\left(A\\right)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nLet <em>A<\/em> = {1, 2, 3, 4, 5, 6} and <em>B<\/em> = {2, 4, 6, 8}.\r\n\r\nWhat is the cardinality of <em>B<\/em>? <em>A<\/em> \u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>?\r\n[reveal-answer q=\"100844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100844\"]\r\n\r\nThe cardinality of <em>B<\/em> is 4, since there are 4 elements in the set.\r\n\r\nThe cardinality of <em>A<\/em> \u22c3<em> B<\/em> is 7, since <em>A<\/em> \u22c3<em> B<\/em> = {1, 2, 3, 4, 5, 6, 8}, which contains 7 elements.\r\n\r\nThe cardinality of <em>A <\/em>\u22c2<em> B<\/em> is 3, since <em>A <\/em>\u22c2<em> B<\/em> = {2, 4, 6}, which contains 3 elements.\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]109846[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nWhat is the cardinality of <em>P<\/em> = the set of English names for the months of the year?\r\n[reveal-answer q=\"6805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"6805\"]\r\n\r\nThe cardinality of this set is 12, since there are 12 months in the year.\r\n\r\nSometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nA survey asks 200 people \u201cWhat beverage do you drink in the morning\u201d, and offers choices:\r\n<ul>\r\n \t<li>Tea only<\/li>\r\n \t<li>Coffee only<\/li>\r\n \t<li>Both coffee and tea<\/li>\r\n<\/ul>\r\nSuppose 20 report tea only, 80 report coffee only, 40 report both.\u00a0\u00a0 How many people drink tea in the morning? How many people drink neither tea or coffee?\r\n[reveal-answer q=\"432276\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"432276\"]\r\n\r\nThis question can most easily be answered by creating a Venn diagram. We can see that we can find the people who drink tea by adding those who drink only tea to those who drink both: 60 people.\r\n\r\nWe can also see that those who drink neither are those not contained in the any of the three other groupings, so we can count those by subtracting from the cardinality of the universal set, 200.\r\n\r\n200 \u2013 20 \u2013 80 \u2013 40 = 60 people who drink neither.\r\n\r\n<img class=\"alignnone size-full wp-image-458\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222016\/coffeetea.png\" alt=\"coffeetea\" width=\"180\" height=\"124\" \/>\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]125872[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA survey asks:\u00a0\u00a0 Which online services have you used in the last month:\r\n<ul>\r\n \t<li>Twitter<\/li>\r\n \t<li>Facebook<\/li>\r\n \t<li>Have used both<\/li>\r\n<\/ul>\r\nThe results show 40% of those surveyed have used Twitter, 70% have used Facebook, and 20% have used both. How many people have used neither Twitter or Facebook?\r\n\r\n[reveal-answer q=\"778879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"778879\"]\r\n\r\nLet <em>T<\/em> be the set of all people who have used Twitter, and <em>F<\/em> be the set of all people who have used Facebook. Notice that while the cardinality of <em>F<\/em> is 70% and the cardinality of <em>T<\/em> is 40%, the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> is not simply 70% + 40%, since that would count those who use both services twice. To find the cardinality of <em>F<\/em> \u22c3 <em>T<\/em>, we can add the cardinality of <em>F<\/em> and the cardinality of <em>T<\/em>, then subtract those in intersection that we\u2019ve counted twice. In symbols,\r\n\r\nn(<em>F<\/em> \u22c3 <em>T<\/em>) = n(<em>F<\/em>) + n(<em>T<\/em>) \u2013 n(<em>F<\/em> \u22c2 <em>T<\/em>)\r\n\r\nn(<em>F<\/em> \u22c3 <em>T<\/em>) = 70% + 40% \u2013 20% = 90%\r\n\r\nNow, to find how many people have not used either service, we\u2019re looking for the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> .\r\n\r\nSince the universal set contains 100% of people and the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> = 90%, the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> must be the other 10%.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>a note about the cardinality properties<\/h3>\r\nYou've already seen how to use the properties of real numbers and how they can be written as \"templates\" or \"forms\" in the general case. The properties of cardinality, although they are not the same as number properties, can be learned in a similar way, by speaking them aloud, writing them out repeatedly, using flashcards, and doing practice problems with them.\r\n\r\nNote below that the first property, spoken aloud, may be expressed as\u00a0<em>the cardinality of set A union with set B will consists of the cardinality of A together with the cardinality of B, after deducting the cardinality of their intersection.\u00a0<\/em>\r\n\r\nThe second property below can be stated as\u00a0<em>the cardinality of the complement of A will consist of the cardinality of the universal set less the cardinality of A.\u00a0<\/em>In other words, it's the cardinality of all the elements that are not in A.\r\n\r\nRemember to employ more than one study strategy along with repetition and practice to learn unfamiliar mathematical concepts.\r\n\r\n<\/div>\r\nThe previous example illustrated two important properties called cardinality properties:\r\n<div class=\"textbox\">\r\n<h3>Cardinality properties<\/h3>\r\n<ol>\r\n \t<li style=\"list-style-type: none;\">\r\n<ol>\r\n \t<li>n(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)<\/li>\r\n \t<li>n(<em>Ac<\/em>) = n(<em>U<\/em>) \u2013 n(<em>A<\/em>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\nNotice that the first property can also be written in an equivalent form by solving for the cardinality of the intersection:\r\n<p style=\"text-align: center;\">n(<em>A<\/em> \u22c2 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c3 <em>B<\/em>)<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>How was that done?<\/h3>\r\nIn the demonstration above, the first cardinality property was rewritten by using the property of equality as you know it from solving equations.\r\n\r\nn(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)\r\n\r\nn(<em>A<\/em> \u22c3 <em>B<\/em>) +\u00a0n(<em>A<\/em> \u22c2 <em>B<\/em>) =\u00a0n(<em>A<\/em>) + n(<em>B<\/em>)\r\n\r\nn(<em>A<\/em> \u22c2 <em>B<\/em>) =\u00a0n(<em>A<\/em>) + n(<em>B<\/em>) \u2013\u00a0n(<em>A<\/em> \u22c3 <em>B<\/em>)\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFifty students were surveyed, and asked if they were taking a social science (SS), humanities (HM) or a natural science (NS) course the next quarter.\r\n\r\n21 were taking a SS course\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 26 were taking a HM course\r\n\r\n19 were taking a NS course\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9 were taking SS and HM\r\n\r\n7 were taking SS and NS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 were taking HM and NS\r\n\r\n3 were taking all three\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7 were taking none\r\n\r\nHow many students are only taking a SS course?\r\n[reveal-answer q=\"88483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"88483\"]\r\n\r\nIt might help to look at a Venn diagram.\r\n\r\n<img class=\"alignnone size-full wp-image-459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222214\/sshmns.png\" alt=\"sshmns\" width=\"180\" height=\"152\" \/>\r\n\r\nFrom the given data, we know that there are\r\n\r\n3 students in region <em>e<\/em> and\r\n\r\n7 students in region <em>h<\/em>.\r\n\r\nSince 7 students were taking a SS and NS course, we know that n(<em>d<\/em>) + n(<em>e<\/em>) = 7. Since we know there are 3 students in region 3, there must be\r\n\r\n7 \u2013 3 = 4 students in region <em>d<\/em>.\r\n\r\nSimilarly, since there are 10 students taking HM and NS, which includes regions <em>e<\/em> and <em>f<\/em>, there must be\r\n\r\n10 \u2013 3 = 7 students in region <em>f<\/em>.\r\n\r\nSince 9 students were taking SS and HM, there must be 9 \u2013 3 = 6 students in region <em>b<\/em>.\r\n\r\nNow, we know that 21 students were taking a SS course. This includes students from regions <em>a, b, d, <\/em>and <em>e<\/em>. Since we know the number of students in all but region <em>a<\/em>, we can determine that 21 \u2013 6 \u2013 4 \u2013 3 = 8 students are in region <em>a<\/em>.\r\n\r\n8 students are taking only a SS course.\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\nOne hundred fifty people were surveyed and asked if they believed in UFOs, ghosts, and Bigfoot.\r\n\r\n43 believed in UFOs\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 44 believed in ghosts\r\n\r\n25 believed in Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 believed in UFOs and ghosts\r\n\r\n8 believed in ghosts and Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5 believed in UFOs and Bigfoot\r\n\r\n2 believed in all three\r\n\r\nHow many people surveyed believed in at least one of these things?\r\n[reveal-answer q=\"252699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252699\"]\r\n\r\nStarting with the intersection of all three circles, we work our way out. Since 10 people believe in UFOs and Ghosts, and 2 believe in all three, that leaves 8 that believe in only UFOs and Ghosts. We work our way out, filling in all the regions. Once we have, we can add up all those regions, getting 91 people in the union of all three sets. This leaves 150 \u2013 91 = 59 who believe in none.\r\n\r\n<img class=\"alignnone size-full wp-image-460\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222338\/ufoghostsbigfoot.png\" alt=\"ufoghostsbigfoot\" width=\"183\" height=\"152\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question]125868[\/ohm_question]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/wErcETeKvrU","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve real-life problems involving sets, subsets, and cardinality properties<\/li>\n<\/ul>\n<\/div>\n<p>Often times we are interested in the number of items in a set or subset. This is called the cardinality of the set.<\/p>\n<div class=\"textbox examples\">\n<h3>about the notation<\/h3>\n<p>In the definition of\u00a0<em>cardinality<\/em> below, note that the symbol [latex]{\\lvert}A{\\rvert}[\/latex] looks like absolute value of [latex]A[\/latex] but does not denote absolute value. This symbol would be understood to represent the cardinality of set [latex]A[\/latex] rather than absolute value by the context in which it is used. Note that the symbol n[latex]\\left(A\\right)[\/latex] is also used to represent the cardinality of set [latex]A[\/latex].<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Cardinality<\/h3>\n<p>The number of elements in a set is the cardinality of that set.<\/p>\n<p>The cardinality of the set <em>A<\/em> is often notated as [latex]{\\lvert}A{\\rvert}[\/latex]\u00a0or\u00a0n[latex]\\left(A\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>Let <em>A<\/em> = {1, 2, 3, 4, 5, 6} and <em>B<\/em> = {2, 4, 6, 8}.<\/p>\n<p>What is the cardinality of <em>B<\/em>? <em>A<\/em> \u22c3<em> B<\/em>, <em>A <\/em>\u22c2<em> B<\/em>?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q100844\">Show Solution<\/span><\/p>\n<div id=\"q100844\" class=\"hidden-answer\" style=\"display: none\">\n<p>The cardinality of <em>B<\/em> is 4, since there are 4 elements in the set.<\/p>\n<p>The cardinality of <em>A<\/em> \u22c3<em> B<\/em> is 7, since <em>A<\/em> \u22c3<em> B<\/em> = {1, 2, 3, 4, 5, 6, 8}, which contains 7 elements.<\/p>\n<p>The cardinality of <em>A <\/em>\u22c2<em> B<\/em> is 3, since <em>A <\/em>\u22c2<em> B<\/em> = {2, 4, 6}, which contains 3 elements.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm109846\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=109846&theme=oea&iframe_resize_id=ohm109846&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>What is the cardinality of <em>P<\/em> = the set of English names for the months of the year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6805\">Show Solution<\/span><\/p>\n<div id=\"q6805\" class=\"hidden-answer\" style=\"display: none\">\n<p>The cardinality of this set is 12, since there are 12 months in the year.<\/p>\n<p>Sometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>A survey asks 200 people \u201cWhat beverage do you drink in the morning\u201d, and offers choices:<\/p>\n<ul>\n<li>Tea only<\/li>\n<li>Coffee only<\/li>\n<li>Both coffee and tea<\/li>\n<\/ul>\n<p>Suppose 20 report tea only, 80 report coffee only, 40 report both.\u00a0\u00a0 How many people drink tea in the morning? How many people drink neither tea or coffee?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q432276\">Show Solution<\/span><\/p>\n<div id=\"q432276\" class=\"hidden-answer\" style=\"display: none\">\n<p>This question can most easily be answered by creating a Venn diagram. We can see that we can find the people who drink tea by adding those who drink only tea to those who drink both: 60 people.<\/p>\n<p>We can also see that those who drink neither are those not contained in the any of the three other groupings, so we can count those by subtracting from the cardinality of the universal set, 200.<\/p>\n<p>200 \u2013 20 \u2013 80 \u2013 40 = 60 people who drink neither.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-458\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222016\/coffeetea.png\" alt=\"coffeetea\" width=\"180\" height=\"124\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm125872\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=125872&theme=oea&iframe_resize_id=ohm125872&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A survey asks:\u00a0\u00a0 Which online services have you used in the last month:<\/p>\n<ul>\n<li>Twitter<\/li>\n<li>Facebook<\/li>\n<li>Have used both<\/li>\n<\/ul>\n<p>The results show 40% of those surveyed have used Twitter, 70% have used Facebook, and 20% have used both. How many people have used neither Twitter or Facebook?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q778879\">Show Solution<\/span><\/p>\n<div id=\"q778879\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>T<\/em> be the set of all people who have used Twitter, and <em>F<\/em> be the set of all people who have used Facebook. Notice that while the cardinality of <em>F<\/em> is 70% and the cardinality of <em>T<\/em> is 40%, the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> is not simply 70% + 40%, since that would count those who use both services twice. To find the cardinality of <em>F<\/em> \u22c3 <em>T<\/em>, we can add the cardinality of <em>F<\/em> and the cardinality of <em>T<\/em>, then subtract those in intersection that we\u2019ve counted twice. In symbols,<\/p>\n<p>n(<em>F<\/em> \u22c3 <em>T<\/em>) = n(<em>F<\/em>) + n(<em>T<\/em>) \u2013 n(<em>F<\/em> \u22c2 <em>T<\/em>)<\/p>\n<p>n(<em>F<\/em> \u22c3 <em>T<\/em>) = 70% + 40% \u2013 20% = 90%<\/p>\n<p>Now, to find how many people have not used either service, we\u2019re looking for the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> .<\/p>\n<p>Since the universal set contains 100% of people and the cardinality of <em>F<\/em> \u22c3 <em>T<\/em> = 90%, the cardinality of (<em>F<\/em> \u22c3 <em>T<\/em>)<em>c<\/em> must be the other 10%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>a note about the cardinality properties<\/h3>\n<p>You&#8217;ve already seen how to use the properties of real numbers and how they can be written as &#8220;templates&#8221; or &#8220;forms&#8221; in the general case. The properties of cardinality, although they are not the same as number properties, can be learned in a similar way, by speaking them aloud, writing them out repeatedly, using flashcards, and doing practice problems with them.<\/p>\n<p>Note below that the first property, spoken aloud, may be expressed as\u00a0<em>the cardinality of set A union with set B will consists of the cardinality of A together with the cardinality of B, after deducting the cardinality of their intersection.\u00a0<\/em><\/p>\n<p>The second property below can be stated as\u00a0<em>the cardinality of the complement of A will consist of the cardinality of the universal set less the cardinality of A.\u00a0<\/em>In other words, it&#8217;s the cardinality of all the elements that are not in A.<\/p>\n<p>Remember to employ more than one study strategy along with repetition and practice to learn unfamiliar mathematical concepts.<\/p>\n<\/div>\n<p>The previous example illustrated two important properties called cardinality properties:<\/p>\n<div class=\"textbox\">\n<h3>Cardinality properties<\/h3>\n<ol>\n<li style=\"list-style-type: none;\">\n<ol>\n<li>n(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)<\/li>\n<li>n(<em>Ac<\/em>) = n(<em>U<\/em>) \u2013 n(<em>A<\/em>)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<p>Notice that the first property can also be written in an equivalent form by solving for the cardinality of the intersection:<\/p>\n<p style=\"text-align: center;\">n(<em>A<\/em> \u22c2 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c3 <em>B<\/em>)<\/p>\n<div class=\"textbox examples\">\n<h3>How was that done?<\/h3>\n<p>In the demonstration above, the first cardinality property was rewritten by using the property of equality as you know it from solving equations.<\/p>\n<p>n(<em>A<\/em> \u22c3 <em>B<\/em>) = n(<em>A<\/em>) + n(<em>B<\/em>) \u2013 n(<em>A<\/em> \u22c2 <em>B<\/em>)<\/p>\n<p>n(<em>A<\/em> \u22c3 <em>B<\/em>) +\u00a0n(<em>A<\/em> \u22c2 <em>B<\/em>) =\u00a0n(<em>A<\/em>) + n(<em>B<\/em>)<\/p>\n<p>n(<em>A<\/em> \u22c2 <em>B<\/em>) =\u00a0n(<em>A<\/em>) + n(<em>B<\/em>) \u2013\u00a0n(<em>A<\/em> \u22c3 <em>B<\/em>)<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Fifty students were surveyed, and asked if they were taking a social science (SS), humanities (HM) or a natural science (NS) course the next quarter.<\/p>\n<p>21 were taking a SS course\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 26 were taking a HM course<\/p>\n<p>19 were taking a NS course\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9 were taking SS and HM<\/p>\n<p>7 were taking SS and NS\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 were taking HM and NS<\/p>\n<p>3 were taking all three\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7 were taking none<\/p>\n<p>How many students are only taking a SS course?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q88483\">Show Solution<\/span><\/p>\n<div id=\"q88483\" class=\"hidden-answer\" style=\"display: none\">\n<p>It might help to look at a Venn diagram.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-459\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222214\/sshmns.png\" alt=\"sshmns\" width=\"180\" height=\"152\" \/><\/p>\n<p>From the given data, we know that there are<\/p>\n<p>3 students in region <em>e<\/em> and<\/p>\n<p>7 students in region <em>h<\/em>.<\/p>\n<p>Since 7 students were taking a SS and NS course, we know that n(<em>d<\/em>) + n(<em>e<\/em>) = 7. Since we know there are 3 students in region 3, there must be<\/p>\n<p>7 \u2013 3 = 4 students in region <em>d<\/em>.<\/p>\n<p>Similarly, since there are 10 students taking HM and NS, which includes regions <em>e<\/em> and <em>f<\/em>, there must be<\/p>\n<p>10 \u2013 3 = 7 students in region <em>f<\/em>.<\/p>\n<p>Since 9 students were taking SS and HM, there must be 9 \u2013 3 = 6 students in region <em>b<\/em>.<\/p>\n<p>Now, we know that 21 students were taking a SS course. This includes students from regions <em>a, b, d, <\/em>and <em>e<\/em>. Since we know the number of students in all but region <em>a<\/em>, we can determine that 21 \u2013 6 \u2013 4 \u2013 3 = 8 students are in region <em>a<\/em>.<\/p>\n<p>8 students are taking only a SS course.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>One hundred fifty people were surveyed and asked if they believed in UFOs, ghosts, and Bigfoot.<\/p>\n<p>43 believed in UFOs\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 44 believed in ghosts<\/p>\n<p>25 believed in Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10 believed in UFOs and ghosts<\/p>\n<p>8 believed in ghosts and Bigfoot\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5 believed in UFOs and Bigfoot<\/p>\n<p>2 believed in all three<\/p>\n<p>How many people surveyed believed in at least one of these things?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q252699\">Show Solution<\/span><\/p>\n<div id=\"q252699\" class=\"hidden-answer\" style=\"display: none\">\n<p>Starting with the intersection of all three circles, we work our way out. Since 10 people believe in UFOs and Ghosts, and 2 believe in all three, that leaves 8 that believe in only UFOs and Ghosts. We work our way out, filling in all the regions. Once we have, we can add up all those regions, getting 91 people in the union of all three sets. This leaves 150 \u2013 91 = 59 who believe in none.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-460\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12222338\/ufoghostsbigfoot.png\" alt=\"ufoghostsbigfoot\" width=\"183\" height=\"152\" \/><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm125868\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=125868&theme=oea&iframe_resize_id=ohm125868&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Sets: cardinality\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/wErcETeKvrU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/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-457\">\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>: Open Textbook Store, Transition Math Project, and the Open Course Library. <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><li>Sets: cardinality. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/wErcETeKvrU\">https:\/\/youtu.be\/wErcETeKvrU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 125872, 109842, 125878. <strong>Authored by<\/strong>:  Bohart,Jenifer, mb Meacham,William. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/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":20,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"Open Textbook 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