{"id":4100,"date":"2013-11-04T09:18:28","date_gmt":"2013-11-04T17:18:28","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=4100"},"modified":"2020-01-15T10:48:52","modified_gmt":"2020-01-15T18:48:52","slug":"difficult-gmat-counting-problems","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/","title":{"rendered":"Difficult GMAT Counting Problems"},"content":{"rendered":"

Here are a few 800+<\/a> counting problems.\u00a0 Yes, that’s right, 800+, which means, as hard as (or possibly harder than) anything you will see on the GMAT.\u00a0 If you can do these, you are in great shape!<\/p>\n

1) How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits equal to 28?<\/p>\n

A. 14<\/p>\n

B. 24<\/p>\n

C. 28<\/p>\n

D. 48<\/p>\n

E. 96<\/p>\n

2) How many distinct four-digit numbers can be formed by the digits {1, 2, 3, 4, 5, 5, 6, 6}?<\/p>\n

A. 280<\/p>\n

B. 360<\/p>\n

C. 486<\/p>\n

D. 560<\/p>\n

E. 606<\/p>\n

\"dcp_img1\"<\/a><\/p>\n

3) In the grid of dots above, each dot has equal vertical & horizontal spacing from the others.\u00a0\u00a0 A small 45-45-90 triangle is drawn.\u00a0 Counting this triangle, how many triangles congruent to this one, of any orientation, can be constructed from dots in this grid?<\/p>\n

A. 112<\/p>\n

B. 120<\/p>\n

C. 240<\/p>\n

D. 448<\/p>\n

E. 480<\/p>\n

4) Fifteen dots are evenly spaced on the circumference of a circle.\u00a0 How many combinations of three dots can we pick from these 15 that do not<\/span><\/i> form an equilateral triangle?<\/p>\n

A. 160<\/p>\n

B. 450<\/p>\n

C. 910<\/p>\n

D. 1360<\/p>\n

E. 2640<\/p>\n

Solutions will come at the end of this article.<\/p>\n

 <\/p>\n

Basic ideas of counting<\/h2>\n

The most basic of all ideas in counting is the Fundamental Counting Principle<\/a>.\u00a0\u00a0 This is more conveniently stated in words than as a formula, so formula-based studiers often overlook the importance of this idea, to their own detriment.\u00a0\u00a0 The more familiar counting ideas — more familiar because they have formulas associated with them!! — are permutations and combinations<\/a>.\u00a0 It’s important to recognize: all the ideas & formulas concerning permutations & combinations can be derived directly from the FCP, and in fact, often in the case of permutations, it’s much simpler to use the FCP directly than to use any permutation formula.<\/p>\n

 <\/p>\n

Not so basic \u2026<\/h2>\n

Those are the basic ideas.\u00a0 They are relatively easy to state, and they are explained in the posts at those links.\u00a0\u00a0 The problem is: when you read an individual counting problem, how do you know which rule to use?<\/p>\n

That’s precisely the hard part of counting, and what’s frustrating is that there’s not a simple flowchart\/procedure we can recommend that will work for all counting problems.\u00a0 The hardest thing about any counting problem is knowing how to begin, and essentially, this a right-brain pattern-matching procedure.\u00a0 In the post How do to GMAT Math Faster<\/a>, I discuss left-brain vs. right brain thinking in greater depth.\u00a0\u00a0 The gist is: any left-brain skill or procedure can be diagrammed and clearly explained in a logical step-by-step way, but any right-brain process is an inherently non-linear process into which one has to grow slowly, through experience.<\/p>\n

Of course, in problem solutions, you can read about how they solved the problem, but this is tricky.\u00a0\u00a0\u00a0 You see, in any counting problem, often the hardest part is how one initially perceives and dissects the situation, the way one breaks it into steps or stages.\u00a0\u00a0 Once that perceptual choice is made, especially if it’s a good choice, then applying any rules or formulas is easy.\u00a0\u00a0 If you just scan the solution for formulas & rules, you will say, yes, when we look it this way, of course we do that calculation<\/i>.\u00a0 \u00a0That’s not what you need to study.\u00a0 Above all, you need to study the very beginning: what was the very choice the solution made in solving the problem?\u00a0 How did they begin the problem?\u00a0 How did they frame the situation?\u00a0 What were their perceptual choices<\/i>?\u00a0\u00a0\u00a0 Left-brain thinkers always want to know what to<\/i> do<\/i>, but in the right-brain realm, the most important choice is how to see<\/i>, and once you are seeing things in the right way, it’s obvious what to do.\u00a0 That’s especially what left-brain thinkers need to get from the solutions to counting problems.\u00a0\u00a0 Patterns emerge over time, and those initial perceptual choices become more and more natural.
\nHere are a few concrete recommendations:<\/p>\n

1) Don’t just memorize the permutation & combinations formulas.\u00a0 Memorization<\/a> is not understanding<\/a>.\u00a0\u00a0 Understand the FCP, and understand how that more fundamental idea underlies the formulas & procedures of permutations & combinations.\u00a0\u00a0 Those are the basic tools you need to know very well.<\/p>\n

2) In counting problems, whether you get the problem right or wrong, always read the solutions.\u00a0 Again, you are looking for how the solution began, the very first perceptual choice made before any calculations were done. \u00a0If you are more of a left-brain person, I actually would recommend keeping notes of the situation given in the question and the first few steps, the perceptual choices, made in the solution.\u00a0\u00a0 Forcing yourself to articulate the choices will help to build connections between the brain’s two hemispheres.<\/p>\n

3) See the How do to GMAT Math Faster<\/a> article for more tips on how to exercise and develop your right-brain.<\/p>\n

 <\/p>\n

Summary<\/h2>\n

If you had trouble with the questions above, it may be that some of the information in this article or at one of the linked articles gives you a clue, but most likely you will have to study the solutions below.\u00a0 Remember to be mindful of what you need to get from the solution to a counting problem, as discussed above.\u00a0 Here’s another counting problem for more practice:<\/p>\n

6. http:\/\/gmat.magoosh.com\/questions\/831<\/a><\/p>\n

If you would like to share your experience of these problems or ask anything about what I’ve written, please leave a comment in the comment section.<\/p>\n

 <\/p>\n

Explanations for practice problems.<\/h2>\n

1) For those familiar with Kakuro<\/a>, this problem presents a related challenge.\u00a0 First, we need to know: how many combinations of four distinct single-digit numbers have a sum of 28?\u00a0 Start with the three highest digits: 9 + 8 + 7 = 24, so {9, 8, 7, 4} is one such combination.\u00a0 First, keep the 9 & 8, and bring the other two closer together: {9, 8, 6, 5}.\u00a0 OK: that exhausts possibilities with 9 & 8 as the highest numbers.\u00a0 Now, omit 8, and try 9 & 7: even with the largest available digits, 9 + 7 + 6 + 5 = 27, so this will not be big enough.\u00a0 Similarly, without 9, even with the largest available digits, 8 + 7 + 6 + 5 = 26, also is not big enough.\u00a0 If we omit either 9 or 8, the sum is not big enough, and with both 9 & 8, there are only two possibilities: {9, 8, 7, 4} or {9, 8, 6, 5}.\u00a0\u00a0 Thus, only two possible sets of four digits.<\/p>\n

For each one of these, we could permutate<\/a> the number in any order.\u00a0 4! = 24, so for each of the two sets, there are 24 different four-digit numbers we could form.\u00a0 That’s a total of 2*24 = 48<\/b> different four-digit numbers.<\/p>\n

Answer = D<\/b><\/p>\n

2) We will break the possible numbers into four different groups.\u00a0 First, we will consider all the numbers with no repeat digits: how many different sets of four different number can I choose from the digits 1-6?\u00a0 Remembering the tricks for calculating combinations<\/a>, 6C4 = 6C2 = 15.\u00a0 For each one of those, we can permutate<\/a> the digits in any order: 4! = 24.\u00a0\u00a0 To multiply 15*24, use the doubling & halving trick<\/a>.\u00a0 Half of 24 is 12, and twice 15 is 30, so 15*24 = 30*12 = 360.\u00a0 That’s group #1, all the four-digit numbers with four distinct digits.<\/p>\n

Group #2 will be the group with two 6’s and the other two digits distinct.\u00a0 Think about this.\u00a0 There are 4C2 = 6 “slots” the two sixes could occupy:<\/p>\n

6 6 _ _<\/p>\n

6 _ 6 _<\/p>\n

6 _ _ 6<\/p>\n

_ 6 6 _<\/p>\n

_ 6 _ 6<\/p>\n

_ _ 6 6<\/p>\n

Now, for each of those, use the FCP<\/a>: we would pick five digits for the first empty slot (1, 2, 3, 4, 5), and that would leave us four choices for the second empty slot.\u00a0 Thus, for each of those six, there are 20 four-digit numbers we could create. That’s 6*20 = 120 total in group #2.<\/p>\n

Group #3 will be the group with two 5’s and the other two digits distinct.\u00a0 Think about this.\u00a0 For symmetry reasons, there must be exactly as many numbers in this group as there are in group #2.\u00a0 There are 120 in this group.<\/p>\n

Group #4 will have two 5’s and two 6’s constituting the four digits.\u00a0\u00a0 If we choose the two slots into which to drop the 5’s, the 6’s have to go into the other slots.\u00a0 Therefore, there must be 4C2 = 6 of these.\u00a0 Here they are: 5566, 5656, 5665, 6556, 6565, 6655.<\/p>\n

The total is the sum across these four groups: 360 + 120 + 120 + 6 = 606<\/b><\/p>\n

Answer = (E)<\/b><\/p>\n

3) First, notice that in every square, we could construct four different triangle.<\/p>\n

\"dcp_img2\"<\/a><\/p>\n

In the above square, we can construct triangles ABC, ACD, ABD, and BCD, all congruent.<\/p>\n

Well, if we take all the squares of this size in the diagram,<\/p>\n

\"dcp_img3\"<\/a><\/p>\n

We have 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 squares.\u00a0 That’s 28 squares, times four triangles in each square: 28*4 = 112.<\/p>\n

BUT, we can’t forget to count the leftover triangles on the diagonal slant:<\/p>\n

\"dcp_img4\"<\/a><\/p>\n

That’s 8 more triangles, for a total of 112 + 8 = 120<\/b><\/p>\n

Answer = (B)<\/b><\/p>\n

4) Well, first of all, ignoring the type of triangle formed, how many combinations total?\u00a0 The easiest way to think about this is to use the Fundamental Counting Principle.\u00a0\u00a0 For the first dot, 15 choices, then 14 left for the second choice, then 13 left for the third choice: that’s 15*14*13.\u00a0 But, that will count repeats: the same three dots could be chosen in any of their 3! = 6 orders, so we have to divide that number by 6.\u00a0 (NOTICE the non-calculator<\/a> math here).<\/p>\n

(15*14*13)\/6<\/p>\n

Cancel the factor of 3 in 15 and 6<\/p>\n

(5*14*13)\/2<\/p>\n

Cancel the factor of 2 in the 14 and 2<\/p>\n

(5*7*13) = 5*91 = 455<\/p>\n

That’s how many total triangles we could create.<\/p>\n

Of these, how many are equilateral triangles?\u00a0 Well, the only equilateral triangles would be three points equally spaced across the whole circle.\u00a0 Suppose the points are numbers from 1 to 15.\u00a0 From point 1 to point 6 is one-third of the circle — again, from point 6 to point 11, and from point 11 back to point 1.\u00a0 That’s one equilateral triangle.\u00a0 We could make an equilateral triangle using points<\/p>\n

{1, 6, 11}<\/p>\n

{2, 7, 12}<\/p>\n

{3, 8, 13}<\/p>\n

(4, 9, 14)<\/p>\n

{5, 10, 15}<\/p>\n

After that, we would start to repeat.\u00a0 There are five possible equilateral triangles, so 455 \u2013 5 = 450<\/b> of these triangles are not equilateral.<\/p>\n

Answer = (B)<\/b><\/p>\n

\"dcp_img5\"<\/a><\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Here are a few 800+ counting problems.\u00a0 Yes, that’s right, 800+, which means, as hard as (or possibly harder than) anything you will see on the GMAT.\u00a0 If you can do these, you are in great shape! 1) How many four digit numbers have no repeat digits, do not contain zero, and have a sum […]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[112],"tags":[],"ppma_author":[13209],"class_list":["post-4100","post","type-post","status-publish","format-standard","hentry","category-math"],"acf":[],"yoast_head":"\nGMAT Counting Problems | Magoosh Study Resources<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Difficult GMAT Counting Problems\" \/>\n<meta property=\"og:description\" content=\"Here are a few 800+ counting problems.\u00a0 Yes, that’s right, 800+, which means, as hard as (or possibly harder than) anything you will see on the GMAT.\u00a0 If you can do these, you are in great shape! 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1) How many four digit numbers have no repeat digits, do not contain zero, and have a sum […]","og_url":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/","og_site_name":"Magoosh Blog \u2014 GMAT\u00ae Exam","article_publisher":"https:\/\/www.facebook.com\/MagooshGMAT\/","article_published_time":"2013-11-04T17:18:28+00:00","article_modified_time":"2020-01-15T18:48:52+00:00","og_image":[{"url":"https:\/\/magoosh.com\/gmat\/files\/2013\/10\/dcp_img1.png"}],"author":"Mike M\u1d9cGarry","twitter_card":"summary_large_image","twitter_creator":"@MagooshGMAT","twitter_site":"@MagooshGMAT","twitter_misc":{"Written by":"Mike M\u1d9cGarry","Est. reading time":"8 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/#article","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/"},"author":{"name":"Mike M\u1d9cGarry","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b"},"headline":"Difficult GMAT Counting Problems","datePublished":"2013-11-04T17:18:28+00:00","mainEntityOfPage":{"@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/"},"wordCount":1570,"commentCount":21,"publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"articleSection":["GMAT Math"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/","url":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/","name":"GMAT Counting Problems | Magoosh Study Resources","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/#website"},"datePublished":"2013-11-04T17:18:28+00:00","breadcrumb":{"@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/magoosh.com\/gmat\/difficult-gmat-counting-problems\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/magoosh.com\/gmat\/"},{"@type":"ListItem","position":2,"name":"Difficult GMAT Counting Problems"}]},{"@type":"WebSite","@id":"https:\/\/magoosh.com\/gmat\/#website","url":"https:\/\/magoosh.com\/gmat\/","name":"Magoosh Blog \u2014 GMAT\u00ae Exam","description":"Everything you need to know about the GMAT","publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/magoosh.com\/gmat\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/magoosh.com\/gmat\/#organization","name":"Magoosh","url":"https:\/\/magoosh.com\/gmat\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/","url":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","contentUrl":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","width":265,"height":60,"caption":"Magoosh"},"image":{"@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/MagooshGMAT\/","https:\/\/twitter.com\/MagooshGMAT"]},{"@type":"Person","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b","name":"Mike M\u1d9cGarry","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/image\/15a1e36ef1c2c3940179212433de141a","url":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","caption":"Mike M\u1d9cGarry"},"description":"Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. 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Learn more about the GMAT through Mike's Youtube video explanations.","sameAs":["https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured"],"award":["Magna cum laude from Harvard"],"knowsAbout":["GMAT"],"knowsLanguage":["English"],"jobTitle":"Content Creator","worksFor":"Magoosh","url":"https:\/\/magoosh.com\/gmat\/author\/mikemcgarry\/"}]}},"authors":[{"term_id":13209,"user_id":26,"is_guest":0,"slug":"mikemcgarry","display_name":"Mike M\u1d9cGarry","avatar_url":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","user_url":"","last_name":"M\u1d9cGarry","first_name":"Mike","description":"Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as \"member of the month\" for over two years at <a href=\"https:\/\/gmatclub.com\/blog\/2012\/09\/mike-mcgarrys-gmat-experience\/\" rel=\"noopener noreferrer\">GMAT Club<\/a>. Mike holds an A.B. in Physics (graduating <em>magna cum laude<\/em>) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's <a href=\"https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured\" rel=\"noopener noreferrer\">Youtube <\/a>video explanations and resources like <a href=\"https:\/\/magoosh.com\/gmat\/whats-a-good-gmat-score\/\" rel=\"noopener noreferrer\">What is a Good GMAT Score?<\/a> and the <a href=\"https:\/\/magoosh.com\/gmat\/gmat-diagnostic-test\/\" rel=\"noopener noreferrer\">GMAT Diagnostic Test<\/a>."}],"_links":{"self":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/4100","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/users\/26"}],"replies":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/comments?post=4100"}],"version-history":[{"count":0,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/4100\/revisions"}],"wp:attachment":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/media?parent=4100"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/categories?post=4100"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/tags?post=4100"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/ppma_author?post=4100"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}