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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Counting Methods ##### This topic has 3 expert replies and 4 member replies ## Counting Methods The five sides of a pentagon have lenghts of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lenghts are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lenghts ? (A) 5 (B) 12 (C) 24 (D) 32 (E) 120 ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15195 messages Followed by: 1861 members Upvotes: 13060 GMAT Score: 790 Top Reply DarkKnight wrote: Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24. The number of ways to arrange n elements around a circular table is (n-1)!. But the number of ways to arrange n elements around a ring is (n-1)!/2. The ring gives us 1/2 the number of distinct arrangements because, unlike a table, a ring can be flipped over. For example: Given the 4 elements ABCD, there are (4-1)! = 3! = 6 ways to arrange them around a circular table. Among these arrangements are ABCD and DBCA. But around a ring, ABCD and DBCA represent the same arrangement, because when the ring is flipped over, ABCD becomes DBCA and DBCA becomes ABCD. So around a ring, the number of distinct arrangements is cut in 1/2, as given in the formula above. Assuming that -- like a ring -- the pentagon in the problem above can be flipped over, the number of distinct ways to arrange the 5 sides = (5-1)!/2 = 12. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Legendary Member Joined 19 Jan 2010 Posted: 586 messages Followed by: 5 members Upvotes: 31 Test Date: 27th July 2011 GMAT Score: 730 Is OA A? Let me know so that I can post the explanation. Dnt want to confuse you more with a wrong explanation if it isnt ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1434 messages Followed by: 32 members Upvotes: 59 koby_gen wrote: The five sides of a pentagon have lenghts of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lenghts are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lenghts ? (A) 5 (B) 12 (C) 24 (D) 32 (E) 120 Hi there! Without loss of generality we may assume that the side of length 6 is positioned (say "horizontally") and we are looking for the number of possible ordered choices (say clockwise chosen) for the lenghts of the other 4 sides (called A, B, C and D in my figure). You have 4 choices for side A (2,3, 4 or 5); this side chosen, you have 3 choices for side B, then 2 choices for side C (and side D is left with the last length among the ones offered). From the Multiplicative Principle, we have 4*3*2 = 24 possible choices. The answer seems to be 24, but it is not. Please note from my second drawings that (according to the question stem) 24 is "duplicated" (I mean: every real distinct configuration was counted twice), therefore the answer is 12. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br Senior | Next Rank: 100 Posts Joined 15 Sep 2008 Posted: 79 messages Upvotes: 3 Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24. Master | Next Rank: 500 Posts Joined 25 Jan 2010 Posted: 406 messages Followed by: 4 members Upvotes: 23 Test Date: March 14, 2014 GMAT Score: 740 Can someone please indicate the source of this question? _________________ I'm really old, but I'll never be too old to become more educated. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15195 messages Followed by: 1861 members Upvotes: 13060 GMAT Score: 790 DarkKnight wrote: Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24. I was asked to explain further why the following is true: (the number of distinct arrangements around a ring) = 1/2*(the number of distinct arrangements around a circular table) Please see the picture below. Let B = blue, G = green, P = purple, R = red. Around a table, the 2 arrangements shown above -- BGPR and BRPG -- represent 2 distinct arrangements. These 2 distinct arrangements will be included when we count that there are (n-1)! = (4-1)! = 6 ways to arrange the four colors around the table. Now imagine rotating the ring around the axis shown in the picture. When the ring is rotated 180 degrees, the arrangement on the left (BGPR) becomes the arrangement on the right (BRPG). Thus, around a ring, each placement of the four elements yields not one but two distinct arrangements. The result is that the number of distinct arrangements is cut in half. The number of ways to arrange the four colors around a ring = (n-1)!/2 = (4-1)!/2 = 3. Does this help? _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. Senior | Next Rank: 100 Posts Joined 15 Sep 2008 Posted: 79 messages Upvotes: 3 GMATGuruNY wrote: DarkKnight wrote: Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24. I was asked to explain further why the following is true: (the number of distinct arrangements around a ring) = 1/2*(the number of distinct arrangements around a circular table) Please see the picture below. Let B = blue, G = green, P = purple, R = red. Around a table, the 2 arrangements shown above -- BGPR and BRPG -- represent 2 distinct arrangements. These 2 distinct arrangements will be included when we count that there are (n-1)! = (4-1)! = 6 ways to arrange the four colors around the table. Now imagine rotating the ring around the axis shown in the picture. When the ring is rotated 180 degrees, the arrangement on the left (BGPR) becomes the arrangement on the right (BRPG). Thus, around a ring, each placement of the four elements yields not one but two distinct arrangements. The result is that the number of distinct arrangements is cut in half. The number of ways to arrange the four colors around a ring = (n-1)!/2 = (4-1)!/2 = 3. Does this help? Thats a neat concept Mitch. Your figures explained it all. Didn't really come across any problem revolving around a "ring" but will certainly keep in mind. Thank you much. Hope you are ready for some more snow in NY. I am dreading already . • Award-winning private GMAT tutoring Register now and save up to$200

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