{"id":13485,"date":"2017-01-12T08:01:14","date_gmt":"2017-01-12T13:01:14","guid":{"rendered":"http:\/\/grockit.com\/gmat\/?p=1138"},"modified":"2020-09-11T20:42:28","modified_gmt":"2020-09-11T20:42:28","slug":"gmat-quantitative-permutations-and-combinations","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/gmat\/gmat-quantitative-permutations-and-combinations\/","title":{"rendered":"GMAT Quantitative: Permutations and Combinations"},"content":{"rendered":"<p>When dealing with permutations and combinations, you are essentially trying to find the number of different outcomes given a set of items and a number of restrictions. The difference between permutation and combination merely depends on whether the order matters. Let me illustrate with an example. Suppose you have three food items, apples, bananas and carrots. If you were asked to only pick two of them, how many possibilities are there? If you got 3, that\u2019s the right answer. You have apples &amp; bananas, apples &amp; carrots and bananas and carrots. This is <strong>combination<\/strong>. If I told you however, that the order in which you eat the food matters, then you have more possibilities, because instead of just apples &amp; bananas, you have to consider bananas &amp; apples too. The latter is <strong>permutation<\/strong>.<br \/>\n&nbsp;<br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h3    avia-builder-el-0  el_before_av_iconlist  avia-builder-el-first  '><h3 class='av-special-heading-tag '  itemprop=\"headline\"  >Listing<\/h3><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nWhat I did earlier, when I listed out the 3 choices is called, not surprisingly, <strong>listing<\/strong>. But there is a method to listing to ensure that you don\u2019t leave any possibilities out. Everyone tends to have his or her own method, let me share mine. If I have four items \u2013 a, b, c and d \u2013 and I\u2019m supposed to choose 2:<br \/>\n<div  class='avia-icon-list-container   avia-builder-el-1  el_after_av_heading  el_before_av_heading '><ul class='avia-icon-list avia-icon-list-left av-iconlist-big avia_animate_when_almost_visible avia-iconlist-animate'>\n<li><div  class='iconlist_icon  avia-font-entypo-fontello'><span class='iconlist-char ' aria-hidden='true' data-av_icon='\ue816' data-av_iconfont='entypo-fontello'><\/span><\/div><article class=\"article-icon-entry \"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\" ><div class='iconlist_content_wrap'><header class=\"entry-content-header\"><h4 class='av_iconlist_title iconlist_title   '  itemprop=\"headline\"  >Step 1<\/h4><\/header><div class='iconlist_content  '  itemprop=\"text\"  ><p>Take the first item <em>a<\/em> and combine it with <em>b. <\/em>Then go down the list and combine it with <em>c<\/em> and <em>d<\/em> to get a total of three possibilities: <em>ab, ac, ad<\/em>.<\/p>\n<\/div><\/div><footer class=\"entry-footer\"><\/footer><\/article><div class='iconlist-timeline'><\/div><\/li>\n<li><div  class='iconlist_icon  avia-font-entypo-fontello'><span class='iconlist-char ' aria-hidden='true' data-av_icon='\ue816' data-av_iconfont='entypo-fontello'><\/span><\/div><article class=\"article-icon-entry \"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\" ><div class='iconlist_content_wrap'><header class=\"entry-content-header\"><h4 class='av_iconlist_title iconlist_title   '  itemprop=\"headline\"  >Step 2<\/h4><\/header><div class='iconlist_content  '  itemprop=\"text\"  ><p>Take the second item <em>b<\/em> and combine it with the rest of the list. Don\u2019t combine it with <em>a<\/em> because you already did that in step 1. So you get: <em>bc, bd<\/em><\/p>\n<\/div><\/div><footer class=\"entry-footer\"><\/footer><\/article><div class='iconlist-timeline'><\/div><\/li>\n<li><div  class='iconlist_icon  avia-font-entypo-fontello'><span class='iconlist-char ' aria-hidden='true' data-av_icon='\ue816' data-av_iconfont='entypo-fontello'><\/span><\/div><article class=\"article-icon-entry \"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\" ><div class='iconlist_content_wrap'><header class=\"entry-content-header\"><h4 class='av_iconlist_title iconlist_title   '  itemprop=\"headline\"  >Step 3<\/h4><\/header><div class='iconlist_content  '  itemprop=\"text\"  ><p>Keep doing the same thing for the next thing on the list, which is <em>c<\/em>. For <em>c<\/em>, there\u2019s only one thing behind it on the list \u2013 <em>d<\/em>. So there\u2019s only 1 possibility here: <em>cd<\/em><\/p>\n<\/div><\/div><footer class=\"entry-footer\"><\/footer><\/article><div class='iconlist-timeline'><\/div><\/li>\n<li><div  class='iconlist_icon  avia-font-entypo-fontello'><span class='iconlist-char ' aria-hidden='true' data-av_icon='\ue816' data-av_iconfont='entypo-fontello'><\/span><\/div><article class=\"article-icon-entry \"  itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/BlogPosting\" itemprop=\"blogPost\" ><div class='iconlist_content_wrap'><header class=\"entry-content-header\"><h4 class='av_iconlist_title iconlist_title   '  itemprop=\"headline\"  >Step 4<\/h4><\/header><div class='iconlist_content  '  itemprop=\"text\"  ><p>Keep going until you get to the second last item. In this case, you\u00a0already reached it in step 3. Tally up your choices: <em>ab, ac, ad, bc, bd, cd<\/em><\/p>\n<\/div><\/div><footer class=\"entry-footer\"><\/footer><\/article><div class='iconlist-timeline'><\/div><\/li>\n<\/ul><\/div><br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h3    avia-builder-el-2  el_after_av_iconlist  el_before_av_image  '><h3 class='av-special-heading-tag '  itemprop=\"headline\"  >Tree Diagram<\/h3><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nIf you noticed, listing is a good method for figuring out combinations. The Tree Diagram is a graphical method that helps you with permutation. Using the previous example of 4 items, here\u2019s the tree diagram that illustrates the number of ways of permuting items.<br \/>\n<div  class='avia-image-container  av-styling-    avia-builder-el-3  el_after_av_heading  el_before_av_heading  avia-align- '  itemprop=\"image\" itemscope=\"itemscope\" itemtype=\"https:\/\/schema.org\/ImageObject\"  ><div class='avia-image-container-inner'><div class='avia-image-overlay-wrap'><img class='wp-image-0 avia-img-lazy-loading-not-0 avia_image' src=\"https:\/\/www.kaptest.com\/blog\/prep\/wp-content\/uploads\/sites\/21\/2017\/01\/3-300x161.jpg\" alt='' title=''   itemprop=\"thumbnailUrl\"  \/><\/div><\/div><\/div><br \/>\nIf you look at the last line of the tree diagram and count the number of boxes, you will see that there are 12 possible ways to permute 2 items out of 4. Each item can be combined with 3 other things. Another way of seeing this is that you have two spaces you need to fill ___ and ___.\u00a0 Looking at the first space, you have 4 possible items you could place there. Looking at the second space, you have 3 possible items you could place there, since you have already put 1 item in the first space. So there are 4 x 3 = 12 ordered possibilities.<br \/>\n&nbsp;<br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h3    avia-builder-el-4  el_after_av_image  el_before_av_promobox  '><h3 class='av-special-heading-tag '  itemprop=\"headline\"  >Fundamental Counting Principle<\/h3><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><br \/>\nThe spaces-and-number-of-options-to-fill-that-space method of thinking is essentially the fundamental counting principle. The principle states that:<br \/>\n\t<div   class='av_promobox  avia-button-no   avia-builder-el-5  el_after_av_heading  el_before_av_heading '>\t\t<div class='avia-promocontent'><p>\n<em><strong>If task A can be done in m ways and after Task A is complete, Task B can be done in n ways, then there are m*n ways of completing Task A then Task B.<\/strong><\/em><\/p>\n<\/div><\/div><br \/>\nLet\u2019s see if you understand the principle with a quick example.<br \/>\n<em>How many different ways are there are arranging these letters: T A N G O.<\/em><br \/>\nDid you get 120?<br \/>\nImagine five spaces ___\u00a0\u00a0 ___\u00a0\u00a0 ___\u00a0\u00a0 ___\u00a0\u00a0 ___<br \/>\nYou have 5 possible letters to put in the first space, and when that\u2019s done, you have 4 possible letters to put in the second space, 3 letters for the third space and so on.<br \/>\nSo the answer is 5*4*3*2*1 or\u00a0120!<br \/>\n&nbsp;<br \/>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h3    avia-builder-el-6  el_after_av_promobox  el_before_av_one_half  '><h3 class='av-special-heading-tag '  itemprop=\"headline\"  >Formulas<\/h3><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><\/p>\n<p class=\"p1\"><span class=\"s1\">Once you\u2019ve figured out whether to use permutation or combination, there is actually very little work to be done if you know the formulas for permutation and combination or if you know where the function is hidden on your calculator.<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">As mentioned before, permutation is used when the order matters and combination when you just want to choose, but not order, the items. Let\u2019s dive straight into the formulas then go through a few examples.<\/span><\/p>\n<p><div class=\"flex_column av_one_half  flex_column_div first  avia-builder-el-7  el_after_av_heading  el_before_av_one_half  \" ><p><strong>Permuting<\/strong><\/p>\n<p class=\"p1\"><span class=\"s1\">The number of ways of permuting r objects out of n objects is given by<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">n <b>P <\/b>r =n!\/(n-r)!\u00a0\u00a0 where n! = n * (n-1) * (n-2) * \u2026 * (2) * (1)<\/span><\/p><\/div><br \/>\n<div class=\"flex_column av_one_half  flex_column_div   avia-builder-el-8  el_after_av_one_half  el_before_av_heading  \" ><p class=\"p1\"><strong>Combining<\/strong><\/p>\n<p class=\"p1\"><span class=\"s1\">The number of ways of combining r objects out of n objects is given by<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">n <b>C<\/b> r = n!\/r!(n-r)!<\/span><\/p><\/div><\/p>\n<p class=\"p1\"><em><span class=\"s1\">Note that 0! is defined as 1.<\/span><\/em><\/p>\n<div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-9  el_after_av_one_half  el_before_av_heading  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Example 1<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div>\n<p class=\"p1\"><span class=\"s1\">How many ways are there of arranging 4 letters out of the following F R I E N D?<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">FRIEND has 6 different letters, and since the order of letters matter, you know you have to use the permutation formula. Applying the formula directly, the answer is given by<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">6 <b>P <\/b>4 = 6!\/(6-4)! &#8211; 6!\/2! &#8211; 360<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">You can double-check this by using the fundamental counting principle that we covered in Part I. Drawing a tree diagram will also work, but it might get a little messy as the tree gets bigger and bigger.<\/span><\/p>\n<p class=\"p1\"><div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-10  el_after_av_heading  el_before_av_heading  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Example 2<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><\/p>\n<p class=\"p1\"><span class=\"s1\">If there are 3 entrees and 5 desserts, how many ways are there of choosing 1 entr\u00e9e and 2 desserts? Note, that the question does not say anything about the order in which the entrees and desserts are eaten, so you\u00a0know to use the combination principle. Because you\u00a0are choosing entrees and desserts separately, you\u00a0have to apply the combination formula twice.<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">First, to choose 1 entr\u00e9e out of 3, we apply the formula 3 <b>C<\/b> 1 = 3!\/1!(3-1)!\u00a0 &#8211; 3!\/1!2! &#8211; 3<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">Next, to choose 2 desserts out of 5, we apply the formula 5 <b>C <\/b>2 = 5!\/2!(5-2)! &#8211; 5!\/2!3! &#8211; 10<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">Then because for each entr\u00e9e, there are 10 possible 2-dessert combinations and there are 3 ways of choosing 1 entr\u00e9e, to get the total number of possibilities, we take 3*10 = 30.<\/span><\/p>\n<p class=\"p1\"><div  style='padding-bottom:10px; ' class='av-special-heading av-special-heading-h4  blockquote modern-quote  avia-builder-el-11  el_after_av_heading  el_before_av_sidebar  '><h4 class='av-special-heading-tag '  itemprop=\"headline\"  >Example 3<\/h4><div class='special-heading-border'><div class='special-heading-inner-border' ><\/div><\/div><\/div><\/p>\n<p class=\"p1\"><span class=\"s1\">What if the question throws you a curveball and asks you to permute something that has a repeated item.\u00a0 For example, how many ways are there of arranging the letters A G H A S T?<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">The \u2018A\u2019 is repeated twice so first you\u00a0pretend that the letters are distinct and find the number of possibilities. Then divide that value by 2! because \u2018A\u2019 is repeated twice.<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">AGHAST has 6 letters, and if we permute all the letters, we get 6!<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">Because A is repeated, we divide 6! by 2! to get the answer 360<\/span><\/p>\n<p class=\"p1\"><span class=\"s1\">Supposing you are asked to permute only 4 out of 6 of the letters from AGHAST, then you would do 6 <b>P <\/b>4 as per normal, and divide that answer by 2! since A is repeated. The answer should be 180.<\/span><\/p>\n<p class=\"p1\"><div  class='avia-builder-widget-area clearfix  avia-builder-el-12  el_after_av_heading  avia-builder-el-last '><div id=\"text-76\" class=\"widget clearfix widget_text\">\t\t\t<div class=\"textwidget\"><p><span data-sumome-listbuilder-embed-id=\"70f62512b89832cc8e7fcd96b9d03245c2794cf0b7170fa257f26acec72436d9\"><\/span><\/p>\n<\/div>\n\t\t<\/div><div id=\"text-77\" class=\"widget clearfix widget_text\">\t\t\t<div class=\"textwidget\"><p><span data-sumome-listbuilder-embed-id=\"71160db829b51adcc3539815988485f372b020407413c2957efe50cfe72ea639\"><\/span><\/p>\n<\/div>\n\t\t<\/div><\/div><\/p>\n","protected":false},"excerpt":{"rendered":"<p>When dealing with permutations and combinations, you are essentially trying to find the number of different outcomes given a set of items and a number of restrictions. The difference between permutation and combination merely depends on whether the order matters. Let me illustrate with an example. Suppose you have three food items, apples, bananas and [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28900,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[55],"tags":[56,80],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13485"}],"collection":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/comments?post=13485"}],"version-history":[{"count":3,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13485\/revisions"}],"predecessor-version":[{"id":36076,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/13485\/revisions\/36076"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media\/28900"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=13485"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=13485"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=13485"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}