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First, try these challenging GMAT Quantitative problems, all variations on a theme, as you will see.<\/p>\nFirst, try these challenging GMAT Quantitative problems, all variations on a theme, as you will see.<\/p>\n
1) Seven children --- A, B, C, D, E, F, and G --- are going to sit in seven chairs in a row.\u00a0 The children C & F have to sit next to each other, and the others can sit in any order in any remaining chairs.\u00a0 How many possible configurations are there for the children?<\/p>\n
2) Seven children --- A, B, C, D, E, F, and G --- are going to sit in seven chairs in a row.\u00a0 Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side.\u00a0 The other four children can sit in any order in any of the remaining seats.\u00a0 How many possible configurations are there for the children?<\/p>\n
3) Six children --- A, B, C, D, E, and F --- are going to sit in six chairs in a row.\u00a0 Child E must be somewhere to the left of child F.\u00a0 How many possible configurations are there for the children?<\/p>\n
4) Seven children --- A, B, C, D, E, F, and G --- are going to sit in seven chairs in a row.\u00a0 Children A & B must sit next to each other, and child C must be somewhere to the right of A & B.\u00a0 How many possible configurations are there for the children?<\/p>\n
If these problems make your head spin, then you have found the right post!<\/p>\n
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First of all, it's vitally important to understand something called the\u00a0Fundamental Counting Principle<\/a>\u00a0(FCP).\u00a0 It's also important to understand\u00a0permutations and combinations<\/a>.\u00a0 These ideas underlie all the introductory counting problems on the GMAT Quantitative section, and they form the basis of what we will discuss in these posts.<\/p>\n <\/p>\n Whenever you are counting arrangements or possibilities in a complex scenario,\u00a0always start with the most restrictive feature or element<\/em><\/strong>.\u00a0 Quite often, the plan is --- (a) count all the possibilities for the elements with restrictions; (b) count all the possibilities for the remaining non-restricted items; (c) by the FCP, multiply those numbers together.<\/p>\n For example, let's take a simple case, a bit simpler than anything the GMAT will throw your way.\u00a0 Suppose we have five children, A, B, C, D, and E, and we are going to sit them in five chairs in a row, with the restriction that B & C have to be next to each other.<\/p>\n Step #1: consider the restricted elements, B & C.\u00a0 They have to be next to each other.\u00a0 How many \"next to each other pairs are there among the four seats?<\/p>\n X X _ _ _ _ _ X X _<\/p>\n _ _ _ X X<\/p>\n There are those four possible pairs.\u00a0 Now, B & C can be in either order in any of those, so that's 4*2 =\u00a08<\/strong>\u00a0possibilities for these two children.<\/p>\n Step #2: now, the non-restricted elements.\u00a0 Once B & C are seated, we can put A & D & E in any order in the remaining three seats.\u00a0 Three elements in any order --- that's a permutation --- 3! = 6.\u00a0 For any particular configuration of B & C, the remaining three can be seated in\u00a06<\/strong>\u00a0different ways.<\/p>\n Step #3: combine with the FCP.\u00a0 Children B & C can be seated in 8 different configurations, and for each one of those, A & D & E can be seated in 6 different configurations.\u00a0 Total number of configurations = 6*8 =\u00a048<\/strong>.<\/p>\n That's the basic process, which can be expanded for larger numbers of items.\u00a0 Having seen that, you may want to give the problems above a second look before reading through the explanations below.\u00a0 If those problems still are confounding for you, read through the solutions below very carefully.\u00a0 If you thoroughly understand the four problems at the top of this post, you will be able to handle anything counting problem GMAT will give you.<\/p>\n <\/p>\n 1) First, we will consider the restricted elements --- C & F.\u00a0 How many ways can they sit next to each other in a row of seven chairs?\u00a0 Well, first of all, how many \"next to each other\" pairs of chairs are there?<\/p>\n X X _ _ _ _ _<\/p>\n _ X X _ _ _ _<\/p>\n _ _ X X _ _ _<\/p>\n _ _ _ X X _ _<\/p>\n _ _ _ _ X X _<\/p>\n _ _ _ _ _ X X<\/p>\n There are six different pairs of \"next to each other\" chairs.\u00a0 For each pair, children C & F could be in either order, so that's 6*2 = 12 possibilities for these two.<\/p>\n Now, consider the other five children.\u00a0 For any configuration of C & F, the remaining five children could be seated in any order among the five remaining seats.\u00a0 Five items in any order --- that's a permutation of the 5 items ---- 5P5 = 5! = 120.\u00a0 For any single configuration of C & F, there are 120 ways that the other children could be seated in the remaining seats.<\/p>\n Finally, we'll combine with the Fundamental Counting Principle.\u00a0 We have 12 ways for the first two, and 120 ways for the remaining five.\u00a0 That's a total number of configurations of 12*120 =\u00a01440<\/strong>.\u00a0 Answer =\u00a0C<\/strong><\/p>\n <\/p>\n 2) First, we will consider the restricted elements --- children A & B & G have to be in three seats in a row.\u00a0 How many \"three in a row\" seats are there in a row of seven seats?<\/p>\n X X X _ _ _ _<\/p>\n _ X X X _ _ _<\/p>\n _ _ X X X _ _<\/p>\n _ _ _ X X X _<\/p>\n _ _ _ _ X X X<\/p>\n There are five different \"three in a row\" locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B --- just those two orders.\u00a0 This means the total number of configurations for these three children is 5*2 = 10.<\/p>\n Now, consider the non-restricted elements, the other four.\u00a0 \u00a0Once A & B & G are seated, the remaining four children can be seated in any order among the remaining four seats\u00a0 --- that's a permutation of the 4 items ---- 4P4 = 4! = 24.\u00a0 For any single configuration of A & B & G, there are 24 ways that the other children could be seated in the remaining seats.<\/p>\n Finally, we'll combine with the Fundamental Counting Principle.\u00a0 We have 10 ways for the first three, and 24 ways for the remaining four.\u00a0 That's a total number of configurations of 24*10 =\u00a0240<\/strong>.\u00a0 Answer =\u00a0A<\/strong><\/p>\n <\/p>\n 3) If we wanted, we could make this one extremely difficult, counting out all kinds of possibilities in several different cases.\u00a0 Instead, we are going to make this ridiculously easy.<\/p>\n First of all, with absolutely no restrictions, how many ways can the six children be arranged on the six chairs?\u00a0 That's a permutation of the 6 items ---- 6P6 = 6! = 720.\u00a0 That's the total number of arrangements with no restrictions.\u00a0 Of course, those 720 arrangements have all kinds of symmetry to them.\u00a0 In particular, in all of those arrangements overall, it's just as likely for E to be to the left of F as it is for E to be to the right of F.\u00a0 Therefore, exactly half must have E to the right of F, and exactly half must have E to the left of F.\u00a0 Therefore, exactly (1\/2)*720 =\u00a0360<\/strong>\u00a0of the arrangements have E to the left of F.\u00a0 Answer =\u00a0D<\/strong>.<\/p>\n <\/p>\n 4) This problem is a notch harder than anything you are likely to see on the GMAT.\u00a0 If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct.<\/p>\n First, consider the restriction of A & B.\u00a0 As with problem #1 above, there are 12 possibilities for A & B, counting both position and order.<\/p>\n Now, put the other five in any order --- that's 120 possibilities, for a total number of configurations of 1440.\u00a0 That number does not take into account the restriction with C.<\/p>\n Think about those 1440 configurations.\u00a0 In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B.\u00a0 Therefore, in (1\/2)*1440 =\u00a0720<\/strong>\u00a0configurations, C will be to the right of A & B.\u00a0 Answer =\u00a0B<\/strong><\/p>\n This post was written by Mike McGarry, GMAT expert at Magoosh<\/a>, and originally posted here<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":" First, try these challenging GMAT Quantitative problems, all variations on a theme, as you will see. 1) Seven children — A, B, C, D, E, F, and G — are…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,736],"tags":[],"class_list":["post-16768","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-quant-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16768","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=16768"}],"version-history":[{"count":1,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16768\/revisions"}],"predecessor-version":[{"id":16770,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/16768\/revisions\/16770"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=16768"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=16768"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=16768"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Start with the restriction<\/h2>\n
\n_ X X _ _<\/p>\nPractice problem explanations<\/h2>\n