<?xml version="1.0" encoding="UTF-8"?><rss version="2.0" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:wfw="http://wellformedweb.org/CommentAPI/" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:sy="http://purl.org/rss/1.0/modules/syndication/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/" > <channel> <title>GMAT Probability and Combinatorics – GMAT</title> <atom:link href="https://www.manhattanprep.com/gmat/blog/tag/gmat-probability-and-combinatorics/feed/" rel="self" type="application/rss+xml" /> <link>https://www.manhattanprep.com/gmat</link> <description>GMAT Prep Course, Best GMAT Class & Study Books | Manhattan Prep GMAT</description> <lastBuildDate>Thu, 05 Sep 2019 15:53:59 +0000</lastBuildDate> <language>en-US</language> <sy:updatePeriod> hourly </sy:updatePeriod> <sy:updateFrequency> 1 </sy:updateFrequency> <generator>https://wordpress.org/?v=6.3.2</generator> <item> <title>Help! I Can’t Handle GMAT Probability and Combinatorics (Part 3)</title> <link>https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-3/</link> <dc:creator><![CDATA[Chelsey Cooley]]></dc:creator> <pubDate>Thu, 13 Apr 2017 16:30:30 +0000</pubDate> <category><![CDATA[Challenge Problem]]></category> <category><![CDATA[For Current Studiers]]></category> <category><![CDATA[GMAT Prep]]></category> <category><![CDATA[GMAT Strategies]]></category> <category><![CDATA[GMAT Study Guide]]></category> <category><![CDATA[Quant]]></category> <category><![CDATA[GMAT Probability and Combinatorics]]></category> <category><![CDATA[Probability and Combinatorics]]></category> <guid isPermaLink="false">http://www.manhattanprep.com/gmat/?p=13515</guid> <description><![CDATA[<p>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re not kidding! Check out our upcoming courses here. In the previous articles in this series, we developed a critical skill for GMAT probability and combinatorics problems: listing out cases. Let’s start by taking […]</p> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-3/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 3)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></description> <content:encoded><![CDATA[<p><img decoding="async" fetchpriority="high" class="alignnone size-full wp-image-13589" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/4-13-17-social-1.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="1200" height="628" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/4-13-17-social-1.png 1200w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/4-13-17-social-1-300x157.png 300w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/4-13-17-social-1-768x402.png 768w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/4-13-17-social-1-1024x536.png 1024w" sizes="(max-width: 1200px) 100vw, 1200px" /></p> <p><b><i>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re not kidding! </i></b><a id="bloglink" href="https://www.manhattanprep.com/gmat/classes/" target="_blank"><b><i>Check out our upcoming courses here</i></b></a><b><i>.</i></b></p> <hr /> <p><b><i></i></b><span style="font-weight: 400;">In the </span><a id="bloglink" href="https://www.manhattanprep.com/gmat/blog/2017/03/30/help-i-cant-handle-gmat-probability-and-combinatorics-part-2/" target="_blank"><span style="font-weight: 400;">previous articles in this series</span></a><span style="font-weight: 400;">, we developed a critical skill for GMAT probability and combinatorics problems: </span><i><span style="font-weight: 400;">listing out cases</span></i><span style="font-weight: 400;">. Let’s start by taking another look at the practice problem from the end of the last article.</span><span id="more-13515"></span></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">Six coworkers (Anil, Boris, Charlie, Dana, Emmaline, and Frank) are having dinner at a restaurant. They’ll 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’t 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 ‘rotating’ the whole table – only focus on the relative positions of the diners.)</span></i></p> <p><span style="font-weight: 400;">You know by now that you should solve problems like this by </span><b>finding an organized way to list the possibilities</b><span style="font-weight: 400;">. The key is to divide the problem into smaller, simpler problems, to make it easier to write that list. </span></p> <p><span style="font-weight: 400;">In this problem, there are only two people who can sit across from Anil. Only Emmaline and Frank will put up with him! We’ll start by looking at</span><b> cases where Emmaline sits across from Anil</b><span style="font-weight: 400;">, and then we’ll look at</span><b> cases where Frank sits across from Anil</b><span style="font-weight: 400;">. Here’s what my scratch work looked like as I started this problem.</span></p> <p><img decoding="async" class="alignnone size-full wp-image-13516" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-1.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="447" height="274" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-1.png 447w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-1-300x184.png 300w" sizes="(max-width: 447px) 100vw, 447px" /></p> <p><span style="font-weight: 400;">Next, I focused in on the table on the left. I know that </span><b>Frank won’t sit next to Emmaline.</b><span style="font-weight: 400;"> So, Frank is either to the left of Anil, or to the right of Anil.</span></p> <p><img decoding="async" class="alignnone size-full wp-image-13517" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-2.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="273" height="159" /></p> <p><span style="font-weight: 400;">Also, </span><b>Dana has to sit next to Emmaline</b><span style="font-weight: 400;">. Combining those two facts together, there are four possible ways to set everything up. Here they are:</span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13518" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-3.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="533" height="138" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-3.png 533w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/gmat-blog-pc-part-3-img-3-300x78.png 300w" sizes="(max-width: 533px) 100vw, 533px" /></p> <p><span style="font-weight: 400;">Finally, Boris and Charlie can sit wherever they’d like. That doubles the number of possibilities, giving us a total of </span><b>8</b><span style="font-weight: 400;">. </span></p> <p><span style="font-weight: 400;">Next, look at the scenarios where Frank sits across from Anil. Try it on your own: can you identify the </span><b>4 possible scenarios</b><span style="font-weight: 400;"> that work? Combining those with the 8 scenarios we found already gives a total of </span><b>12 ways to arrange the diners</b><span style="font-weight: 400;">. </span></p> <p><span style="font-weight: 400;">Now, that practice problem is more time-consuming than anything you’re likely to see on the GMAT. However, you can use the basic ideas from that problem on a wide range of GMAT probability and combinatorics problems! </span></p> <p><span style="font-weight: 400;">On the GMAT, you may have heard about “order matters” and “order doesn’t matter” combinatorics problems—let’s talk a bit about those. Personally, I find the “order matters” terminology to be confusing, so I’m not going to use those words in this article. Instead, here’s a method that will use the skill of “splitting up the problem” that you’ve already been developing. Here’s an example problem.</span></p> <p style="padding-left: 30px;"><em><span style="font-weight: 400;">In how many ways can a committee of 5 members be chosen from a class of 10 people? </span></em></p> <p style="padding-left: 30px;"><span style="font-weight: 400;">(A) 40<br /> </span><span style="font-weight: 400;">(B) 126<br /> </span><span style="font-weight: 400;">(C) 252<br /> </span><span style="font-weight: 400;">(D) 6048<br /> </span><span style="font-weight: 400;">(E) 30240</span></p> <p><span style="font-weight: 400;">Yikes—those numbers are too big to </span><i><span style="font-weight: 400;">just</span></i><span style="font-weight: 400;"> list out the possibilities, like we have been. How do we start? Just like we have been, start by making the problem simpler. Don’t try to do the whole thing at once. Instead of asking yourself how to choose the committee, ask yourself: </span><b>in how many ways can I choose the first member of the committee?</b><span style="font-weight: 400;"> Well, there are 10 people in the class, so there are 10 ways to do it. </span></p> <p><span style="font-weight: 400;">Now, zoom in. If you’ve already chosen the first member, how many ways are there to choose the second member? There are 9 people left, so there are 9 ways to choose. For each of 10 first members, there are 9 second members. Keep doing this until you’ve chosen the entire committee: (10)(9)(8)(7)(6).</span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13519" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/cc-42.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="650" height="37" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/cc-42.png 650w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/cc-42-300x17.png 300w" sizes="(max-width: 650px) 100vw, 650px" /></p> <p>One of my high school math teachers loved to tell this joke:</p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">How many legs does a cow have?</span></i></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">Eight: two front legs, two back legs, two left legs, and two right legs. </span></i></p> <p><span style="font-weight: 400;">When we found the solution above, we made the exact same mistake. We counted each committee more than once, just like my math teacher counted each leg more than once. Let’s see how it happened. </span></p> <p><span style="font-weight: 400;">We started by picking the first committee member. Let’s say that was Anil. Then, we picked a second committee member: that was Boris. Then, we picked Charlie, then Dana, then Emmaline. However, we </span><i><span style="font-weight: 400;">separately </span></i><span style="font-weight: 400;">counted the scenario where we started by picking Dana, then picked Anil, then Charlie, then Emmaline, then Boris. And we </span><i><span style="font-weight: 400;">also</span></i><span style="font-weight: 400;"> separately counted the case where we picked Anil, then Charlie, then Boris, then Dana, then Emmaline. Those committees should </span><i><span style="font-weight: 400;">actually</span></i><span style="font-weight: 400;"> all be the </span><i><span style="font-weight: 400;">same </span></i><span style="font-weight: 400;">committee, because they have the same people on them! We should have only counted it once. But we messed up—we counted it a bunch of different times, just like we counted each of the cow’s legs too many times. </span></p> <p><span style="font-weight: 400;">However, you can fix this problem easily. You just have to </span><b>divide by the number of times you counted each committee.</b><span style="font-weight: 400;"> (Notice that in my math teacher’s joke, we counted each leg twice—so, to get the right answer, you’d just have to divide by 2). How many times did we count the committee consisting of Anil, Boris, Charlie, Dana, and Emmaline? We counted it once for each possible order we could have picked them in. There are (5)(4)(3)(2)(1) different orders to put them in, which means we counted each committee 120 times. Our answer is 120 times too big. </span></p> <p><span style="font-weight: 400;">Just divide by 120, and you have the right answer! Here’s what your scratch work would look like:</span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13520" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/04/cc-42-2.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 3) by Chelsey Cooley" width="136" height="44" /></p> <p><span style="font-weight: 400;">Okay, try it again, but more quickly: </span></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">How many different poker hands of 5 cards can be drawn from a 52-card deck?</span></i></p> <p><span style="font-weight: 400;">There are 52 ways to draw the first card, 51 ways to draw the second card, and so on. That makes (52)(51)(50)(49)(48) different hands. </span><b>Did we overcount?</b><span style="font-weight: 400;"> Yes! We counted each hand (5)(4)(3)(2)(1) = 120 times, just like we counted each committee 120 times. Divide by 120 to get your answer. </span></p> <p><span style="font-weight: 400;">How about this one? </span></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">A teacher asks 3 students from a class of 7 to stand in a straight line from left to right. In how many different ways can this be done? </span></i></p> <p><span style="font-weight: 400;">Okay, there are 7 ways to choose the student on the left, then 6 students who could be in the middle, then 5 students on the right. (7)(6)(5) possibilities. </span><b>Did we overcount? </b><span style="font-weight: 400;">Actually, we didn’t. We don’t have to divide. That’s because the line consisting of Anil, Boris, and Charlie </span><i><span style="font-weight: 400;">should</span></i><span style="font-weight: 400;"> be counted separately from the line consisting of Charlie, Boris, and Anil. Those are two different lines, so you need to count them both individually. The right answer is (7)(6)(5) = 210—no division necessary. </span></p> <p><span style="font-weight: 400;">Now you have a basic strategy for combinatorics problems with </span><b>bigger numbers</b><span style="font-weight: 400;">: count the possibilities as simply as possible. Then, </span><b>divide if necessary</b><span style="font-weight: 400;"> to fix the “cow has eight legs” mistake. This strategy achieves the same thing as thinking about “order matters,” but it keeps you from having to worry about which type of problem you’re dealing with! Instead, you can just use math and common sense to figure it out. ?</span></p> <hr /> <p><b><i>See that “SUBSCRIBE” button in the top right corner? Click on it to receive all our GMAT blog updates straight to your inbox!</i></b></p> <hr /> <p><b><i><em><strong><a id="bloglink" href="https://www.manhattanprep.com/instructors/chelsey-cooley/" target="_blank">Chelsey Cooley</a><a href="https://www.manhattanprep.com/instructors/chelsey-cooley/?utm_source=manhattanprep.com%2Fgre%2Fblog&utm_medium=blog&utm_content=CooleyBioGREBlog&utm_campaign=GRE%20Blog" target="_blank"><img decoding="async" loading="lazy" class="alignleft" title="Chelsey Cooley Manhattan Prep GRE Instructor" src="https://cdn2.manhattanprep.com/gre/wp-content/uploads/sites/19/2015/11/chelsey-cooley-150x150.jpg" alt="Chelsey Cooley Manhattan Prep GRE Instructor" width="150" height="150" /></a> is a Manhattan Prep instructor based in Seattle, Washington.</strong> </em></i></b><i><em>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’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE. </em></i><i><em><a id="bloglink" href="https://www.manhattanprep.com/gre/classes/#instructor/48" target="_blank">Check out Chelsey’s upcoming GRE prep offerings here</a>.</em></i></p> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-3/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 3)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></content:encoded> </item> <item> <title>Help! I Can’t Handle GMAT Probability and Combinatorics (Part 2)</title> <link>https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-2/</link> <dc:creator><![CDATA[Chelsey Cooley]]></dc:creator> <pubDate>Thu, 30 Mar 2017 15:42:48 +0000</pubDate> <category><![CDATA[Challenge Problem]]></category> <category><![CDATA[For Current Studiers]]></category> <category><![CDATA[GMAT Prep]]></category> <category><![CDATA[GMAT Strategies]]></category> <category><![CDATA[GMAT Study Guide]]></category> <category><![CDATA[How to Study]]></category> <category><![CDATA[Quant]]></category> <category><![CDATA[GMAT Probability]]></category> <category><![CDATA[GMAT Probability and Combinatorics]]></category> <guid isPermaLink="false">http://www.manhattanprep.com/gmat/?p=13430</guid> <description><![CDATA[<p>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re 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> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-2/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 2)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></description> <content:encoded><![CDATA[<p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13509" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-30-17-social-1.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 2) by Chelsey Cooley" width="1200" height="628" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-30-17-social-1.png 1200w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-30-17-social-1-300x157.png 300w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-30-17-social-1-768x402.png 768w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-30-17-social-1-1024x536.png 1024w" sizes="(max-width: 1200px) 100vw, 1200px" /></p> <p><b><i>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re not kidding! </i></b><a id="bloglink" href="https://www.manhattanprep.com/gmat/classes/" target="_blank"><b><i>Check out our upcoming courses here</i></b></a><b><i>.</i></b></p> <hr /> <p><b><i></i></b><span style="font-weight: 400;">In the </span><a id="bloglink" href="https://www.manhattanprep.com/gmat/blog/2017/03/16/help-i-cant-handle-gmat-probability-and-combinatorics-part-1/" target="_blank"><span style="font-weight: 400;">previous article in this series</span></a><span style="font-weight: 400;">, we introduced two big ideas about GMAT probability and combinatorics:</span></p> <ol> <li style="font-weight: 400;"><span style="font-weight: 400;">Most people find them counterintuitive.</span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">The best way to get past that is to </span><i><span style="font-weight: 400;">list the possibilities.</span></i></li> </ol> <p><span style="font-weight: 400;">In this article, we’ll focus more on #2. How do you list out the possibilities in a GMAT probability or combinatorics problem? Let’s try it on a simple probability problem.</span><span id="more-13430"></span></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">If you flip three fair coins, what is the probability of getting exactly two heads? </span></i></p> <p><span style="font-weight: 400;">Student A reasons as follows: “There 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 ¼.”</span></p> <p><span style="font-weight: 400;">Unfortunately, Student A is incorrect. That’s because </span><i><span style="font-weight: 400;">Student A’s four scenarios aren’t equally likely</span></i><span style="font-weight: 400;">. It’s actually more likely that you’ll 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> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13431" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41.png" alt="" width="654" height="189" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41.png 654w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-300x87.png 300w" sizes="(max-width: 654px) 100vw, 654px" /></p> <p><span style="font-weight: 400;">In </span><b>three out of the eight cases</b><span style="font-weight: 400;">, there are exactly two heads. The probability is 3/8.</span></p> <p><span style="font-weight: 400;">How can you be sure that you’ve considered every case? The trick is to stay organized. To create the table above, I thought like this: </span></p> <ul> <li style="font-weight: 400;"><span style="font-weight: 400;">Let’s start with three heads. There’s only one way for that to happen. List it out.</span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">Now, two heads. In that case, there’s one tail. The one tail could be the first coin, the second coin, or the third coin. List out those cases.</span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">Next, one head. That single head could be the first coin, the second coin, or the third coin. That’s three more cases. List them out.</span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">Finally, there could be zero heads. That’s the last case.</span></li> </ul> <p><span style="font-weight: 400;">I’m sure that I listed every case, because I considered all of the possibilities, from ‘all heads’ down to ‘no heads.’ But actually, that’s not the only way to list your cases! Here’s a second way to get the same result. </span></p> <p><span style="font-weight: 400;">This time, start by listing out the cases that start with two heads. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13432" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-2.png" alt="" width="655" height="49" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-2.png 655w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-2-300x22.png 300w" sizes="(max-width: 655px) 100vw, 655px" /></p> <p>Then, consider the cases that start with two tails:</p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13433" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-3.png" alt="" width="655" height="49" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-3.png 655w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-3-300x22.png 300w" sizes="(max-width: 655px) 100vw, 655px" /></p> <p>Then, consider the cases that start with a head and a tail:</p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13434" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-4.png" alt="" width="659" height="48" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-4.png 659w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-4-300x22.png 300w" sizes="(max-width: 659px) 100vw, 659px" /></p> <p>Finally, the cases that start with a tail and a head:</p> <p><span style="font-weight: 400;"><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13435" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-5.png" alt="" width="654" height="45" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-5.png 654w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-5-300x21.png 300w" sizes="(max-width: 654px) 100vw, 654px" /></span></p> <p><span style="font-weight: 400;">There are still eight cases, and three of them still have exactly two heads. </span><b>Once again, we can conclude that the probability is 3/8.</b></p> <p><span style="font-weight: 400;">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> <p style="padding-left: 30px;"><i><span style="font-weight: 400;"> A family of five is going on a road trip. Their car has five seats: a driver’s 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’t sit in the back, because she gets carsick, and Alice and Claire can’t sit next to each other, or else they’ll fight. In how many different ways can the family sit in the car? </span></i></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13436" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-6.png" alt="" width="661" height="37" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-6.png 661w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-6-300x17.png 300w" sizes="(max-width: 661px) 100vw, 661px" /></p> <p>Got your list? Okay, I’ll share how I thought about this one.</p> <p><span style="font-weight: 400;">Let’s 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> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13437" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-7.png" alt="" width="658" height="52" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-7.png 658w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-7-300x24.png 300w" sizes="(max-width: 658px) 100vw, 658px" /></p> <p>If Dad is driving, then Mom must be in the front passenger seat. She can’t be in the back, and that’s the only seat left. That leaves the three kids in the back.</p> <p><span style="font-weight: 400;">If we put the three kids in the back, how many ways can we arrange them? Well, Bob has to be in the middle – otherwise, Alice and Claire would be next to each other, and we can’t 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’s the other way around. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13438" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-8.png" alt="" width="656" height="70" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-8.png 656w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-8-300x32.png 300w" sizes="(max-width: 656px) 100vw, 656px" /></p> <p>There are two possibilities, and they look like this:</p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13439" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/gmat-blog-pc-part-2-img-1.png" alt="" width="630" height="255" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/gmat-blog-pc-part-2-img-1.png 630w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/gmat-blog-pc-part-2-img-1-300x121.png 300w" sizes="(max-width: 630px) 100vw, 630px" /></p> <p><b>What have we just learned? </b><span style="font-weight: 400;">First, if Dad is driving, there are only two different ways to seat everyone else. Second, despite my GMAT score, I’m not so great at drawing cars. </span></p> <p><span style="font-weight: 400;"> Okay, let’s turn our attention to the second scenario. What if Mom is driving?</span></p> <p><span style="font-weight: 400;">Anybody could be in the front seat, so let’s split it up again to make things simpler. Let’s suppose that Dad is in the front seat. How many ways could we arrange everybody else? </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13440" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-10.png" alt="" width="653" height="71" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-10.png 653w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-10-300x33.png 300w" sizes="(max-width: 653px) 100vw, 653px" /></p> <p>If Dad is in the front with Mom, we have the same two cases we already looked at. Alice and Claire can’t be next to each other, so Bob is in the middle and his sisters are on either side.</p> <p><span style="font-weight: 400;">What if Dad </span><i><span style="font-weight: 400;">isn’t</span></i><span style="font-weight: 400;"> in the front? Let’s try putting poor Bob in the front, instead of sticking him in the middle seat. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13441" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-11.png" alt="" width="655" height="127" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-11.png 655w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-11-300x58.png 300w" sizes="(max-width: 655px) 100vw, 655px" /></p> <p>Dad would have to be in the middle, to separate Alice and Claire. That gives us two more possibilities.</p> <p><span style="font-weight: 400;">What if Alice is in the front? Well, it doesn’t matter how we arrange the back seat, since there’s no risk of Alice and Claire fighting. There are 6 ways to arrange it, and they all work fine. </span><b>The same thing happens if we put Claire in front</b><span style="font-weight: 400;">: since one of the sisters is in the front, it doesn’t matter what order the back seat is in. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13442" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-12.png" alt="" width="663" height="247" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-12.png 663w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/cc-41-12-300x112.png 300w" sizes="(max-width: 663px) 100vw, 663px" /></p> <p>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> <p><span style="font-weight: 400;">Here’s what I really want you to notice. </span><b>This process isn’t magic, and it isn’t math. </b><span style="font-weight: 400;">It’s just about organization. </span></p> <ul> <li style="font-weight: 400;"><span style="font-weight: 400;">First, we </span><b>split up the problem</b><span style="font-weight: 400;"> to make it easier. This is one of the basic principles of counting cases. </span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">We split it up that way for a good reason. </span><b>There were only two different people who could drive</b><span style="font-weight: 400;">, so splitting up the problem like that made it easy to stay organized. If we’d split up the problem based on who was in the passenger seat (for example), we still would’ve gotten the right answer, but it would have been much messier. </span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">When we realized that Mom driving still gave us a lot of different cases, we </span><b>split up the problem a second time</b><span style="font-weight: 400;">. </span></li> <li style="font-weight: 400;"><span style="font-weight: 400;">We stayed organized by </span><b>drawing a chart on our paper and filling it in</b><span style="font-weight: 400;">. </span></li> </ul> <p><span style="font-weight: 400;">This problem only had a small number of scenarios. There’s one big question remaining: can we use the same technique to handle much larger problems? We can. In two weeks, <a id="bloglink" href="https://www.manhattanprep.com/gmat/blog/2017/04/13/help-i-cant-handle-gmat-probability-and-combinatorics-part-3/" target="_blank">check out the final article in this series</a> to learn how. And while you’re waiting, try this practice problem: </span></p> <p style="padding-left: 30px;"><i><span style="font-weight: 400;">Six coworkers (Anil, Boris, Charlie, Dana, Emmaline, and Frank) are having dinner at a restaurant. They’ll 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’t 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 ‘rotating’ the whole table – only focus on the relative positions of the diners.) </span></i><span style="font-weight: 400;">?</span></p> <hr /> <p><b><i>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’re not kidding. </i></b><a id="bloglink" href="https://www.manhattanprep.com/gmat/classes/" target="_blank"><b><i>Check out our upcoming courses here</i></b></a><b><i>.</i></b><span style="font-weight: 400;"> </span></p> <hr /> <p><b><i><em><strong><a id="bloglink" href="https://www.manhattanprep.com/instructors/chelsey-cooley/" target="_blank">Chelsey Cooley</a><a href="https://www.manhattanprep.com/instructors/chelsey-cooley/?utm_source=manhattanprep.com%2Fgre%2Fblog&utm_medium=blog&utm_content=CooleyBioGREBlog&utm_campaign=GRE%20Blog" target="_blank"><img decoding="async" loading="lazy" class="alignleft" title="Chelsey Cooley Manhattan Prep GRE Instructor" src="https://cdn2.manhattanprep.com/gre/wp-content/uploads/sites/19/2015/11/chelsey-cooley-150x150.jpg" alt="Chelsey Cooley Manhattan Prep GRE Instructor" width="150" height="150" /></a> is a Manhattan Prep instructor based in Seattle, Washington.</strong> </em></i></b><i><em>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’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE. </em></i><i><em><a id="bloglink" href="https://www.manhattanprep.com/gre/classes/#instructor/48" target="_blank">Check out Chelsey’s upcoming GRE prep offerings here</a>.</em></i></p> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-2/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 2)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></content:encoded> </item> <item> <title>Help! I Can’t Handle GMAT Probability and Combinatorics (Part 1)</title> <link>https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-1/</link> <dc:creator><![CDATA[Chelsey Cooley]]></dc:creator> <pubDate>Thu, 16 Mar 2017 15:57:23 +0000</pubDate> <category><![CDATA[Challenge Problem]]></category> <category><![CDATA[For Current Studiers]]></category> <category><![CDATA[GMAT Prep]]></category> <category><![CDATA[GMAT Strategies]]></category> <category><![CDATA[GMAT Study Guide]]></category> <category><![CDATA[How to Study]]></category> <category><![CDATA[Quant]]></category> <category><![CDATA[Word Problems]]></category> <category><![CDATA[GMAT Probability and Combinatorics]]></category> <category><![CDATA[Probability and Combinatorics]]></category> <guid isPermaLink="false">http://www.manhattanprep.com/gmat/?p=13314</guid> <description><![CDATA[<p>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re not kidding! Check out our upcoming courses here. There’s a classic brain teaser called the Monty Hall problem. It’s named after the host of an old-timey TV game show, who used it to […]</p> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-1/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 1)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></description> <content:encoded><![CDATA[<p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13415" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-16-17-social-1.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 1) by Chelsey Cooley" width="1200" height="628" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-16-17-social-1.png 1200w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-16-17-social-1-300x157.png 300w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-16-17-social-1-768x402.png 768w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-16-17-social-1-1024x536.png 1024w" sizes="(max-width: 1200px) 100vw, 1200px" /></p> <p><b><i>Did you know that you can attend the first session of any of our online or in-person GMAT courses absolutely free? We’re not kidding! </i></b><a id="bloglink" href="https://www.manhattanprep.com/gmat/classes/" target="_blank"><b><i>Check out our upcoming courses here</i></b></a><b><i>.</i></b></p> <hr /> <p><b><i></i></b><span style="font-weight: 400;">There’s a classic brain teaser called the Monty Hall problem. It’s named after the host of an old-timey TV game show, who used it to confound contestants. He’d present each contestant with three closed doors. Behind one door was a new car, and behind the other two doors were goats. </span></p> <p><span style="font-weight: 400;">Monty invited the player to pick one of the three doors. Whichever door the player chose, Monty would then open a </span><i><span style="font-weight: 400;">different</span></i><span style="font-weight: 400;"> one, revealing a goat, not the car. Then, he would offer the player a choice. If the player wanted, he could </span><i><span style="font-weight: 400;">switch</span></i><span style="font-weight: 400;"> doors, picking the other unopened door. Or, he could stick with the door he picked in the first place. Whichever decision he made, he would win the prize behind the door he chose. </span><span id="more-13314"></span></p> <p><span style="font-weight: 400;">The question is, </span><i><span style="font-weight: 400;">does switching doors give the contestant a higher probability of picking the car</span></i><span style="font-weight: 400;">? It should seem obvious that it doesn’t. The car is equally likely to be behind any of the doors. So, it seems like once one of the three doors is open, you have 50/50 odds of picking the car, regardless of whether you switch or stay. </span></p> <p><span style="font-weight: 400;">However, </span><i><span style="font-weight: 400;">that isn’t true</span></i><span style="font-weight: 400;">. Believe it or not, switching actually increases the probability of picking the car to 2/3, while staying on the same door means that you only have a 1 in 3 chance of picking it. If you didn’t figure that out on your own, you’re in the company of many very smart people, including a number of famous mathematicians.</span></p> <p><span style="font-weight: 400;">From a GMAT perspective, here’s what to take away from this little anecdote. GMAT probability and combinatorics are counterintuitive for </span><i><span style="font-weight: 400;">everyone</span></i><span style="font-weight: 400;">. I majored in math in college – I even took a couple of courses on probability – and I still can’t solve these problems by intuiting which formulas to use. It seems crazy to me that switching doors after the fact would make you more likely to pick the car. So, what do I do – and what should you do? I think about almost all GMAT probability and combinatorics problems in terms of </span><i><span style="font-weight: 400;">counting out possibilities</span></i><span style="font-weight: 400;">. </span></p> <p><span style="font-weight: 400;">Let’s give it a try with the Monty Hall problem. Imagine that you play three games with Monty. Additionally, imagine we’re in probability land, where everything happens exactly according to its probability. Since there are three possible scenarios, and they’re all equally likely, here’s what your three games will look like. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13316" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/1-cc.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 1) by Chelsey Cooley" width="673" height="108" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/1-cc.png 673w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/1-cc-300x48.png 300w" sizes="(max-width: 673px) 100vw, 673px" /></p> <p><span style="font-weight: 400;">Now, Monty opens one of the two doors you didn’t pick. Remember that Monty always opens a door with a goat behind it. So, in two out of the three games, Monty is stuck: there’s only one door he can choose. </span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13317" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/2-cc.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 1) by Chelsey Cooley" width="668" height="99" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/2-cc.png 668w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/2-cc-300x44.png 300w" sizes="(max-width: 668px) 100vw, 668px" /></p> <p><span style="font-weight: 400;">In both of these games, you started with a goat. If you switch, you’ll win! </span></p> <p><span style="font-weight: 400;">In game 3, Monty could pick either door, since they both have goats. Let’s say he flips a coin to decide which one to pick. If he gets heads, he’ll open the first mystery door. If he gets tails, he’ll open the second one. This scenario only comes up when you picked the car in the first place – so, Monty only needs to worry about it 1/3 of the time.</span></p> <p><img decoding="async" loading="lazy" class="alignnone size-full wp-image-13318" src="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-cc.png" alt="Manhattan Prep GMAT Blog - Help! I Can't Handle GMAT Probability and Combinatorics (Part 1) by Chelsey Cooley" width="664" height="98" srcset="https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-cc.png 664w, https://cdn2.manhattanprep.com/gmat/wp-content/uploads/sites/18/2017/03/3-cc-300x44.png 300w" sizes="(max-width: 664px) 100vw, 664px" /></p> <p><span style="font-weight: 400;">Interestingly, it doesn’t matter what Monty flips. No matter what, if Monty flips a coin at all, and you switch, you’ll lose. If you always switch, you’ll win 2 out of your 3 games, and lose 1 of them. That means that switching gives you a 2/3 probability of winning. </span></p> <p><span style="font-weight: 400;">The reason the Monty Hall problem is so counterintuitive, is that people don’t remember that you’re more likely to start with a goat than with a car. Since there are two possibilities – goat or car – your brain wants to think that they’re equally likely. They aren’t. If you start with a goat, switching is always good, and if you start with a car, switching is always bad. You’re more likely to start with a goat than with a car, so without any more information, you should assume that switching is good. </span></p> <p><span style="font-weight: 400;">How do you apply this to the GMAT? There are a few ways. First, recognize that probability is counterintuitive. Not just for you – for many people who are far better at math than you and I, as well. If you see a super tough GMAT probability and combinatorics problem, feel free to guess, since everyone else is probably getting it wrong too. And </span><i><span style="font-weight: 400;">don’t</span></i><span style="font-weight: 400;"> make the “obvious” guess, since it’s probably a trap. Second, the best way to work through counterintuitive GMAT probability and combinatorics problems is to write out, or think out, the different possibilities in an organized way. This is even possible when you’re doing problems with very large numbers, as you’ll learn from <a id="bloglink" href="https://www.manhattanprep.com/gmat/blog/2017/03/30/help-i-cant-handle-gmat-probability-and-combinatorics-part-2/" target="_blank">the next article in this series</a>. ?</span></p> <hr /> <p><b><i>Want more guidance from our GRE gurus? You can attend the first session of any of our online or in-person GRE courses absolutely free! We’re not kidding. </i></b><a id="bloglink" href="https://www.manhattanprep.com/gre/classes/" target="_blank"><b><i>Check out our upcoming courses here</i></b></a><b><i>.</i></b></p> <hr /> <p><b><i><em><strong><a id="bloglink" href="https://www.manhattanprep.com/instructors/chelsey-cooley/" target="_blank">Chelsey Cooley</a><a href="https://www.manhattanprep.com/instructors/chelsey-cooley/?utm_source=manhattanprep.com%2Fgre%2Fblog&utm_medium=blog&utm_content=CooleyBioGREBlog&utm_campaign=GRE%20Blog" target="_blank"><img decoding="async" loading="lazy" class="alignleft" title="Chelsey Cooley Manhattan Prep GRE Instructor" src="https://cdn2.manhattanprep.com/gre/wp-content/uploads/sites/19/2015/11/chelsey-cooley-150x150.jpg" alt="Chelsey Cooley Manhattan Prep GRE Instructor" width="150" height="150" /></a> is a Manhattan Prep instructor based in Seattle, Washington.</strong> </em></i></b><i><em>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’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE. </em></i><i><em><a id="bloglink" href="https://www.manhattanprep.com/gre/classes/#instructor/48" target="_blank">Check out Chelsey’s upcoming GRE prep offerings here</a>.</em></i></p> <p>The post <a rel="nofollow" href="https://www.manhattanprep.com/gmat/blog/help-i-cant-handle-gmat-probability-and-combinatorics-part-1/">Help! I Can’t Handle GMAT Probability and Combinatorics (Part 1)</a> appeared first on <a rel="nofollow" href="https://www.manhattanprep.com/gmat">GMAT</a>.</p> ]]></content:encoded> </item> </channel> </rss>