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Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We\u2019re not kidding! <\/i><\/b>Check out our upcoming courses here<\/i><\/b><\/a>.<\/i><\/b><\/p>\n <\/i><\/b>In the <\/span>previous article in this series<\/span><\/a>, we introduced two big ideas about GMAT probability and combinatorics:<\/span><\/p>\n In this article, we\u2019ll focus more on #2. How do you list out the possibilities in a GMAT probability or combinatorics problem? Let\u2019s try it on a simple probability problem.<\/span><\/p>\n If you flip three fair coins, what is the probability of getting exactly two heads? <\/span><\/i><\/p>\n Student A reasons as follows: \u201cThere are four different scenarios. You could get no heads, one head, two heads, or three heads. Therefore, the probability of getting two heads is one out of four, or \u00bc.\u201d<\/span><\/p>\n Unfortunately, Student A is incorrect. That\u2019s because <\/span>Student A\u2019s four scenarios aren\u2019t equally likely<\/span><\/i>. It\u2019s actually more likely that you\u2019ll get two heads than three heads. To see why (and to solve this problem), break down the possibilities more finely. Actually write out which coins will land heads up, and which ones will land tails up.<\/span><\/p>\n In <\/span>three out of the eight cases<\/b>, there are exactly two heads. The probability is 3\/8.<\/span><\/p>\n How can you be sure that you\u2019ve considered every case? The trick is to stay organized. To create the table above, I thought like this: <\/span><\/p>\n I\u2019m sure that I listed every case, because I considered all of the possibilities, from \u2018all heads\u2019 down to \u2018no heads.\u2019 But actually, that\u2019s not the only way to list your cases! Here\u2019s a second way to get the same result. <\/span><\/p>\n This time, start by listing out the cases that start with two heads. <\/span><\/p>\n Then, consider the cases that start with two tails:<\/p>\n Then, consider the cases that start with a head and a tail:<\/p>\n Finally, the cases that start with a tail and a head:<\/p>\n There are still eight cases, and three of them still have exactly two heads. <\/span>Once again, we can conclude that the probability is 3\/8.<\/b><\/p>\n Before you keep reading, try listing out all of the scenarios that fit the following description. Stay organized, and use visuals if they help you!<\/span><\/p>\n A family of five is going on a road trip. Their car has five seats: a driver\u2019s seat and one passenger seat in the front, and two window seats and a middle seat in the back. The family consists of Mom, Dad, and three kids: Alice, Bob, and Claire. Only Mom and Dad are able to drive. Mom can\u2019t sit in the back, because she gets carsick, and Alice and Claire can\u2019t sit next to each other, or else they\u2019ll fight. In how many different ways can the family sit in the car? <\/span><\/i><\/p>\n Got your list? Okay, I\u2019ll share how I thought about this one.<\/p>\n Let\u2019s start with the driver, since that position has the fewest possibilities. Either Mom or Dad is driving. So, the first thing I jotted down on my paper was a two-column chart:<\/span><\/p>\n If Dad is driving, then Mom must be in the front passenger seat. She can\u2019t be in the back, and that\u2019s the only seat left. That leaves the three kids in the back.<\/p>\n If we put the three kids in the back, how many ways can we arrange them? Well, Bob has to be in the middle \u2013 otherwise, Alice and Claire would be next to each other, and we can\u2019t have that. So there are really only two ways to do it. Either Alice is on the left and Claire is on the right, or it\u2019s the other way around. <\/span><\/p>\n There are two possibilities, and they look like this:<\/p>\n What have we just learned? <\/b>First, if Dad is driving, there are only two different ways to seat everyone else. Second, despite my GMAT score, I\u2019m not so great at drawing cars. \u00a0<\/span><\/p>\n Okay, let\u2019s turn our attention to the second scenario. What if Mom is driving?<\/span><\/p>\n Anybody could be in the front seat, so let\u2019s split it up again to make things simpler. Let\u2019s suppose that Dad is in the front seat. How many ways could we arrange everybody else? <\/span><\/p>\n If Dad is in the front with Mom, we have the same two cases we already looked at. Alice and Claire can\u2019t be next to each other, so Bob is in the middle and his sisters are on either side.<\/p>\n What if Dad <\/span>isn\u2019t<\/span><\/i> in the front? Let\u2019s try putting poor Bob in the front, instead of sticking him in the middle seat. <\/span><\/p>\n Dad would have to be in the middle, to separate Alice and Claire. That gives us two more possibilities.<\/p>\n What if Alice is in the front? Well, it doesn\u2019t matter how we arrange the back seat, since there\u2019s no risk of Alice and Claire fighting. There are 6 ways to arrange it, and they all work fine. <\/span>The same thing happens if we put Claire in front<\/b>: since one of the sisters is in the front, it doesn\u2019t matter what order the back seat is in. <\/span><\/p>\n If Mom is driving, there are 2 + 2 + 6 + 6 = 16 different ways for the family to sit. Add the two cases where Dad is driving, and that makes a total of 18 possibilities.<\/p>\n Here\u2019s what I really want you to notice. <\/span>This process isn\u2019t magic, and it isn\u2019t math. <\/b>It\u2019s just about organization. <\/span><\/p>\n This problem only had a small number of scenarios. There\u2019s one big question remaining: can we use the same technique to handle much larger problems? We can. In two weeks, check out the final article in this series<\/a> to learn how. And while you\u2019re waiting, try this practice problem: \u00a0<\/span><\/p>\n Six coworkers (Anil, Boris, Charlie, Dana, Emmaline, and Frank) are having dinner at a restaurant. They\u2019ll sit in chairs that are evenly spaced around a circular table. Boris, Charlie, and Dana refuse to sit directly across from Anil, because he chews with his mouth open. Frank and Emmaline won\u2019t sit next to each other. Finally, Emmaline and Dana insist on sitting next to each other. How many different arrangements will work? (Ignore the arrangements that come from \u2018rotating\u2019 the whole table \u2013 only focus on the relative positions of the diners.) <\/span><\/i>?<\/span><\/p>\n Want more guidance from our GMAT gurus? You can attend the first session of any of our online or in-person GMAT courses absolutely free! We\u2019re not kidding. <\/i><\/b>Check out our upcoming courses here<\/i><\/b><\/a>.<\/i><\/b>\f<\/span><\/p>\n Chelsey Cooley<\/a> Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We\u2019re not kidding! Check out our upcoming courses here. In the previous article in this series, we introduced two big ideas about GMAT probability and combinatorics: Most people find them counterintuitive. The best way […]<\/p>\n","protected":false},"author":127,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13,873,929,52871,930,2,8],"tags":[53007,53000],"yst_prominent_words":[],"class_list":["post-13430","post","type-post","status-publish","format-standard","hentry","category-challenge-problem","category-for-current-studiers","category-gmat-prep","category-gmat-strategies","category-gmat-study-guide","category-how-to-study","category-quant-on-gmat","tag-gmat-probability","tag-gmat-probability-and-combinatorics"],"_links":{"self":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/13430","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/users\/127"}],"replies":[{"embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/comments?post=13430"}],"version-history":[{"count":7,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/13430\/revisions"}],"predecessor-version":[{"id":13604,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/posts\/13430\/revisions\/13604"}],"wp:attachment":[{"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/media?parent=13430"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/categories?post=13430"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/tags?post=13430"},{"taxonomy":"yst_prominent_words","embeddable":true,"href":"https:\/\/www.manhattanprep.com\/gmat\/wp-json\/wp\/v2\/yst_prominent_words?post=13430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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\n![\"Chelsey](\"https:\/\/cdn2.manhattanprep.com\/gre\/wp-content\/uploads\/sites\/19\/2015\/11\/chelsey-cooley-150x150.jpg\" "\\"Chelsey")
<\/a> is a Manhattan Prep instructor based in Seattle, Washington.<\/strong>\u00a0<\/em><\/i><\/b>Chelsey always followed her heart when it came to her education. Luckily, her heart led her straight to the perfect background for GMAT and GRE teaching: she has undergraduate degrees in mathematics and history, a master\u2019s degree in linguistics, a 790 on the GMAT, and a perfect 170\/170 on the GRE.\u00a0<\/em><\/i>Check out Chelsey\u2019s upcoming GRE prep offerings here<\/a>.<\/em><\/i><\/p>\n","protected":false},"excerpt":{"rendered":"