Permutations with Restrictions
Posted on
02
Mar 2021

Permutations with Restrictions

By: Rich Zwelling (Apex GMAT Instructor)
Date: 2 March 2021

So far, we’ve covered the basics of GMAT combinatorics, the difference between permutations and combinations, some basic permutation and combination math, and permutations with repeat elements. Now, we’ll see what happens when permutation problems involve conceptual restrictions that can obscure how to approach the math.

To illustrate this directly, let’s take a look at the following Official Guide problem:

The letters D, G, I, I , and T can be used to form 5-letter strings as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A) 12
B) 18
C) 24
D) 36
E) 48

Did you catch the restriction? Up until the end, this is a standard permutation with repeats combinatorics problem, since there are five letters and two repeats of the letter ‘I’. However, we’re suddenly told that the two I’s must be separated by at least one other letter. Put differently, they are not allowed to be adjacent.

So how do we handle this? Well, in many cases, it’s helpful to set aside what we want and instead consider what we don’t want. It seems counterintuitive at first, but if we consider the number of ways in which the two I’s can appear together (i.e. what is not allowed) and then subtract that number from the total number of permutations without any restrictions, wouldn’t we then be left with the number of ways in which the two I’s would not appear together (i.e. what is allowed)? 

Let’s demonstrate: 

In this case, we’ll pretend this problem has no restrictions. In the word “DIGIT,” there are five letters and two I’s. Using the principle discussed in our Permutations with Restrictions post, this would produce 5! / 2! = 60 permutations. 

However, we now want to subtract out the permutations that involve the two I’s side by side, since this condition is prohibited by the problem. This is where things become less about math and more about logic and conceptual understanding. Situationally, how would I outline every possible way the two I’s could be adjacent? Well, if I imagine the two I’s grouped together as one unit, there are four possible ways for this to happen:

II DGT

D II GT

DG II T

DGT II

For each one of these four situations, however, the three remaining letters can be arranged in 3*2*1 = 6 ways. 

That produces a total of 6*4 = 24 permutations in which the two I’s appear side by side.

Subtract that from the original 60, and we have: 60 – 24 = 36. The correct answer is D

As you can see, this is not about a formula or rote memorization but instead about logic and analytical skills. This is why tougher combinatorics questions are more likely to involve restrictions.

Here’s another Official Guide example. As always, give it a shot before reading on:

Of the 3-digit integers greater than 700, how many have 2 digits that are equal to each other and the remaining digit different from the other 2 ?

(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

Explanation

This is a classic example of a problem that will tie you up in knots if you try to brute force it. You could try writing up examples that fit the description, such as 717, 882, 939, or 772, trying to find some kind of pattern based on what does work. But as with the previous problem, what if we examine conceptually what doesn’t work?

This will be very akin to how we handle some GMAT probability questions. The situation desired is 2 digits equal and 1 different. What other situations are there (i.e. the ones not desired)?  Well, if you take a little time to think about it, there are only two other possibilities: 

  1. The digits are all the same
  2. The digits are all different

If we can figure out the total number of permutations without restrictions and subtract out the number of permutations in the two situations just listed, we will have our answer. 

First, let’s get the total number of permutations without restrictions. In this case, that’s just all the numbers from 701 up to 999. (Be careful of the language. Since it says “greater than 700”, we will not include 700.)

To get the total number of terms, we must subtract the two numbers then add one to account for the end point. So there are (999-701)+1 = 299 numbers in total without restrictions.

(Another way to see this is that the range between 701 and 999 is the same as the range between 001 and 299, since we simply subtracted 700 from each number, keeping the range identical. It’s much easier to see that there are 299 numbers in the latter case.)

Now for the restrictions. How many of these permutations involve all the digits being the same? Well, this is straightforward enough to brute force: there are only 3 cases, namely 777, 888, and 999. 

How about all the digits being different? Here’s where we have to use our blank (or slot) method for each digit:

___ ___ ___

How many choices do we have for the first digit? The only choices we have are 7, 8, and 9. That’s three choices:

_3_  ___ ___

Once that first digit is in place, how many choices do we have left for the second slot? Well, there are 10 digits, but we have to remove the one already used in the first slot from consideration, as every digit must be different. That means we have nine left:

_3_  _9_  ___

Using the same logic, that leaves us eight for the final slot:

_3_  _9_  _8_

Multiplying them together, we have 3*9*8 = 216 permutations in which the digits are different.

So there are 216+3 = 219 restrictions, or permutations that we do not want. We can now subtract that from the total of 299 total permutations without restrictions to get our final answer of 299-219 = 80. The correct answer is C.

Next time, we’ll take a look at a few examples of combinatorics problems involving COMBINATIONS with restrictions.

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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5 Takeaways from a Successful EA Journey
Posted on
25
Feb 2021

5 Takeaways from a Successful EA Journey

By: Apex GMAT
Contributor: Uerda Muca
Date: 25 February 2021

Each client that contacts us is in a different stage of their Executive Assessment prep, but a universal constant is that each is striving for a great 165+ EA score. The threshold difficulty standing in their way is the lack of a proper mindset, which in turn can lead to a poor performance, whether due attitude, inefficient solving mechanisms, misplaced focus, or myriad other issues. No matter what, mindset leads the way to performance.

To adjust one’s way of perceiving problems requires much more intricate work than cramming a bunch of material, facts, and figures. Taking the time to understand this and elevate your approach to the test is challenging but ultimately rewarding come test day. Here are five takeaways that anyone scoring 150 or better on the Executive Assessment comes to realize along their EA journey. These insights that help test takers thrive help top performers continue to excel in their EMBA/grad school programs and in their post-EMBA careers, long after the Executive Assessment is a distant memory. 

It is not what you know… it’s what you do with it. 

The EA is a psychometric exam. It expects you to be knowledgeable in a core group of secondary school concepts. It’s not a knowledge test, but it uses this universe of information as a baseline that everyone reasonably has been exposed to long before they thought about the Executive Assessment. The exam tests not so much your knowledge but your creative application of that knowledge.

In the process of preparing for your EA it is vital to maximize your performance, which necessitates deep understanding of seemingly straightforward concepts so that you can be flexible in how you navigate them. For instance, in the Integrated Reasoning section there is a high chance that you will come across an unfamiliar graph you need to use. In such a case the ability to draw conclusions from known graphs and apply them to the new situation is much more valuable than having seen the specific graph before.

This holds true well beyond the exam. The amount of information you will be exposed to within the 2 years of a top tier EMBA program is staggering. In order to thrive in this demanding environment you must be selective, actively deciding what information you take on to master, and use universal thinking tools (heuristics and mental models) to be adaptable as new concepts and information come your way.

For the Executive Assessment, the core concepts are indeed essential. But it is also important to notice what concepts and information you can derive from fundamental knowledge and how to do so, hence not needing to memorize it. Knowing how to successfully apply your knowledge will result in efficiency which will afford you the ability and time to excel in the EA, explore what your EMBA has to offer, and be a thought leader in your chosen career. 

Prioritization is crucial 

On the Executive Assessment there are harsh penalties for unanswered questions, so it is vital to complete each section in the time allotted. Therefore, proper time and process management is critical when sitting the exam. Essentially, each problem represents a decision where you must weigh the likelihood of obtaining a correct answer, your time commitment to that problem, ancillary considerations like stress and focus management, and how this problem fits into your larger strategy for the section and the exam. Ultimately, you must decide how much time it is worth expending on each problem as part of your core process.

This mental cost benefit analysis must be deeply embedded in your thought process to achieve an elite EA score. With the proper calibration, this sense will certainly be useful in business school and beyond. In the professional world, there will always be time constraints – be it stringent deadlines or time zone differences. Being able to prioritize focus and make decisions quickly and accurately while navigating uncertainty and incomplete information is a huge strength. Similarly, actively choosing to abandon a low value or less important task so that you can fully devote to solving an issue of importance is not a sign of weakness or incapability, but rather an asset in a world that will always ask more of you than you can give. Time is scarce in the workplace, and just like on the Executive Assessment, you should prioritize what adds the most value to your bottom line. 

Every problem has multiple solution paths

 A common theme in our client’s feedback is their fascination with a core principle that we teach; that every EA problem has multiple solution paths and that sensitivity to how you solve the problem is more important than simply arriving at the correct answer. Let’s take a means and averages problem from the Quant section as an example. Many would be tempted to solve this mathematically straightaway, but this problem can be solved more efficiently using a scenario or a graph rather than processing equations, delivering greater clarity and freeing up valuable time for other, more challenging problems.

Wresting yourself away from the paradigm that a problem has a single “correct” solution path is essential to conquering the Executive Assessment but is also valuable in life. Very few things are clear cut and unambiguous, and training yourself to recognize multiple ways to get to the same destination is important, especially if you can recognize them before committing to any specific path. Seeking answers beyond the ordinary and obvious will provide you with innovative ways of overcoming obstacles and drive progress, and make you a thought leader among your peers and in your graduate program and organization.

Focusing on the structure of the EA helps you compare solution paths and choose the best for the current challenge, resulting in not only a correct answer, but a timely one. Moreover, thinking of a problem from multiple perspectives means that you take into consideration unlikely or unnoticed features of a problem, and when applied to a business setting, this added vision can drive great insight into stakeholders interests and uncover innovative solutions to intractable problems.

In order to succeed first know yourself

The Executive Assessment is not an exam where you can get 100% of the problems correct. In fact, your score will not depend on the number of questions that you get correct, but rather by the difficulty level of the ones that you do get correct. Since the EA is computer adaptive, it increases in difficulty until it matches a candidate’s capabilities, and the aim as a test taker is to get to the most difficult problems that you can handle, and then get most of those correct. In this way, the Executive Asssessment drives you to perform at your best rather than spending a lot of time testing fundamentals.

Ultimately, this means that you must decide how to allocate your time and energy to produce the best performance. This means understanding your strengths and weaknesses, evaluating each problem in light of those, and then deciding which problems make sense to handle, which make sense to invest extra time in, and which (few) problems you might want to walk away from right off the bat in order to preserve your valuable time for higher value problems. Don’t simply put your head down and try to get everything right – at least not at first.

During your EA preparation you should be conscious of how you perform and how you have progressed from where you began. If you struggle to finish a practice exam in a timely manner, this is a sign that your time management skills require polishing, and that you’re not making timing decisions well. If you perform well on Sentence Correction but not on the Reading Comprehension, then that means that you can spend less time on SC questions and reallocate that time towards Reading Comp, thereby increasing your score and building confidence in your Executive Assessment allocation decisions along the way.

Understanding your strengths and weaknesses, and the workload you handle best will help you excel in your career. Furthermore, knowing your capabilities can aid you when setting work boundaries and defining your professional skill set on the other side of business school. Successful professionals know how to focus on what they do best, and remove those tasks that impinge upon their productivity and value.

In this way, they don’t find themselves taking on too much, and are able to have work life balance, all while placing them in a position to continue to achieve because of that balance. Overworking is counterproductive because it drives burnout and reduces focus and efficiency. In much the same way that athletes require proper rest for peak performance, those working in intellectually rigorous fields requiring creativity need mental breaks for better focus, clarity and job performance. In this sense, being aware of your own limitations will guide you towards a healthy work-life balance and in turn increase productivity. 

5% talent and 95% hard work

Being naturally intuitive with numbers or extremely well-read can provide a great footing for your EA preparation. Without further development, however, natural talent can only take you so far. The Executive Assessment begins testing you the moment you can no longer trust your intuition and talent, and then need to rely upon knowing what you don’t know, and navigating towards deeper insights. The EA tests a range of skills such as critical assessment of data, ability to reason and analytical thinking. This means that being knowledgeable and skillful with fundamentals, or being a strong student only lays the foundation for success. It’s persistence, determination, and having a comprehensive study plan and clear understanding of this exam’s architecture that defines those who score 165 or better.

The good news is that the skills necessary to get a 165+ EA score can be cultivated and enhanced with hard work, perseverance and determination. Moreover, these same skills can help you get the most out of your EMBA program and career and enhance your skill set. For example, in business school you may come across an exceptional mathematician pursuing a concentration in Marketing because she has identified a weak point, and wants to focus on how to conduct research, to write and communicate clearly and effectively and to understand and implement data in the decision making process. Similarly, someone with average mathematical Mathematical ability might excel in Finance courses because of the skills he has developed – analytical thinking, problem solving, and constructing mental models.

Conclusion

The most important thing is to put in the hard work (effortful learning, not just a lot of prep time) to grow those top-level skills, regardless of how naturally gifted you are in a given subject. Marketing isn’t all creativity nor is finance all math, and in this way professional challenges are similar to the Executive Assessment itself, which is neither about Math nor English grammar. Compensating your weaknesses and enhancing your strengths in your chosen concentration will be a vital part of your EMBA experience, and it should start with your EA preparation.

Your Executive Assessment journey can be pleasant and enriching rather than an arduous, distasteful experience that you dread having to go through. With the proper mindset, guidance and support you can grow through your EA experience to acquire valuable skills that will help you for years to come.

Schedule a call with one of our experienced Executive Assessment consultants at +41 41 534 98 78, +44 (0) 79 4361 2406 or +1 (917) 819-5945  and get a head start on the road to achieving your goal.

If you enjoyed this EA journey article, learn more about the EA exam in this overview.

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What happens when Permutations have repeat elements?
Posted on
25
Feb 2021

What happens when Permutations have repeat elements?

By: Rich Zwelling (Apex GMAT Instructor)
Date: 25 February 2021

Permutations With Repeat Elements

As promised in the last post, today we’ll discuss what happens when we have a PERMUTATIONS situation with repeat elements. What does this mean exactly? Well, let’s return to the basic example in our intro post on GMAT combinatorics:

If we have five distinct paintings, and we want to know how many arrangements can be created from those five, we simply use the factorial to find the answer (i.e. 5! = 5*4*3*2*1 = 120). Let’s say those paintings were labeled A, B, C, D, and E. 

Now, each re-arrangement of those five is a different PERMUTATION, because the order is different:

ABCDE
EBADC
CADBE


etc

Remember, there are 120 permutations because if we use the blank (or slot) method, we would have five choices for the first blank, and once that painting is in place, there would be four left for the second blank, etc…

_5_  _4_  _3_  _2_  _1_ 

…and we would multiply these results to get 5! or 120.

However, what if, say we suddenly changed the situation such that some of the paintings were identical? Let’s replace painting C with another B and E with another D:

ABBDD

Suddenly, the number of permutations decreases, because some paintings are no longer distinct. And believe it or not, there’s a formulaic way to handle the exact number of permutations. All you have to do is take the original factorial, and divide it by the factorials of each repeat. In this case, we have 5! for our original five elements, and we now must divide by 2! for the two B’s and another 2! for the two D’s:

  5!
——
2! 2!     

= 5*4*3*2*1
   ————-
  (2*1)(2*1)

= 5*2*3
= 30 permutations

As another example, try to figure out how many permutations you can make out of the letters in the word BOOKKEEPER? Give it a shot before reading the next paragraph.

In the case of BOOKKEEPER, there are 10 letters total, so we start with a base of 10! 

We then have two O’s, two K’s and three E’s for repeats, so our math will look like this:

   10!
———
2! 2! 3! 

Definitely don’t calculate this, though, as GMAT math stays simple and likes to come clean. Remember, we’ll have to divide out the repeats. You are extremely unlikely to have to do this calculation for a GMAT problem, however, since it relies heavily on busy-work mechanics. The correct answer choice would thus look like the term above. 

Let’s now take a look at an Official Guide question in which this principle has practical use. I’ll leave it to you to discover how. As usual, give the problem a shot before reading on:

A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

(A) 3/8
(B) 1/4
(C) 3/16
(D) 1/8
(E) 1/16

Quick Probability Review

Remember from our post of GMAT Probability that, no matter how complicated the problem, probability always boils down to the basic concept of:

    Desired Outcomes
———————————–
Total Possible Outcomes

In this case, each child has two equally likely outcomes: boy and girl. And since there are four children, we can use are blank method to realize that we’ll be multiplying two 4 times:

_2_  _2_  _2_  _2_   =  16 total possible outcomes (denominator)

This may give you the premature notion that C or E must be correct, simply because you see a 16 in the denominator, but remember, fractions can reduce! We could have 4 in the numerator, giving us a fraction of 4/16, which would reduce to 1/4. And every denominator in the answer choices contains a factor of 16, so we can’t eliminate any answers based on this. 

Now, for the Desired Outcomes component, we must figure out how many outcomes consist of exactly two boys and two girls. The trick here is to recognize that it could be in any order. You could have the two girls followed by the two boys, vice versa, or have them interspersed. Now, you could brute-force this and simply try writing out every possibility. However, you must be accurate, and there’s a chance you’ll forget some examples. 

What if we instead write out an example as GGBB for two girls and two boys? Does this look familiar? Well, this should recall PERMUATIONS, as we are looking for every possible ordering in which the couple could have two girls and two boys. And yes, we have two G’s and two B’s as repeats. Here’s the perfect opportunity to put our principle into play:

We have four children, so we use 4! for our numerator, then we divide by 2! twice for each repeat:

  4!
——
2! 2! 

This math is much simpler, as the numerator is 24, while the denominator is 4. (Remember, memorize those factorials up to 6!)

This yields 6 desired outcomes of two boys and two girls. 

With 6 desired outcomes of 16 total possible outcomes, our final probability fraction is 6/16, which reduces to 3/8. The correct answer is A.

Next time, we’ll look into combinatorics problems that involve restrictions, which can present interesting conceptual challenges. 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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An Introduction to combination math
Posted on
23
Feb 2021

An Intro to Combination Math

By: Rich Zwelling (Apex GMAT Instructor)
Date: 23 Feb 2021

Last time, we looked at the following GMAT combinatorics practice problem, which gives itself away as a PERMUTATION problem because it’s concerned with “orderings,” and thus we care about the order in which items appear:

At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge?

A) 120
B) 240
C) 360
D) 480
E) 600

(Review the previous post if you’d like an explanation of the answer.)

Now, let’s see how a slight frame change switches this to a COMBINATION problem:

At a farmers market, a chef is to sell some of his best cheeses. Due to the market’s very bizarre restrictions, he can sell exactly two or three cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential groupings of cheeses can he create for display to customers? 

A) 6
B) 15
C)
20
D) 35
E) 120

Did you catch why this is a COMBINATION problem instead of a PERMUTATION problem? The problem asked about “groupings.” This implies that we care only about the items involved, not the sequence in which they appear. Cheddar followed by brie followed by gouda is not considered distinct from brie followed by gouda followed by cheddar, because the same three cheeses are involved, thus producing the same grouping

So how does the math work? Well, it turns out there’s a quick combinatorics formula you can use, and it looks like this: 

combinations problem

Let’s demystify it. The left side is simply notational, with the ‘C’ standing for “combination.” The ‘n’ and the ‘k’ indicate larger and smaller groups, respectively. So if I have a group of 10 paintings, and I want to know how many groups of 4 I can create, that would mean n=10 and k=4. Notationally, that would look like this:

combinatorics and permutations on the GMAT, combination math on the gmat

Now remember, the exclamation point indicates a factorial. As a simple example, 4! = 4*3*2*1. You simply multiply every positive integer from the one given with the factorial down to one. 

So, how does this work for our problem? Let’s take a look:

At a farmers market, a chef is to sell some of his best cheeses. Due to the market’s very bizarre restrictions, he can sell exactly two or three cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential groupings of cheeses can he create for display to customers? 

A) 6
B) 15
C)
20
D) 35
E) 120

The process of considering the two cases independently will remain the same. It cannot be both two and three cheeses. So let’s examine the two-cheese case first. There are six cheese to choose from, and we are choosing a subgroup of two. That means n=6 and k=2:

combinations and permutation on the gmat, combination math on the gmat

Now, let’s actually dig in and do the math:

combinatorics and permutations on the GMAT, combination math on the gmat

combinatorics and permutations on the GMAT, combination math on the gmat

From here, you’ll notice that 4*3*2*1 cancels from top and bottom, leaving you with 6*5 = 30 in the numerator and 2*1 in the denominator:

combinatorics and permutations on the GMAT, combination math on the gmat That leaves us with:

6C2 = 15 combinations of two cheeses

Now, how about the three-cheese case? Similarly, there are six cheeses to choose from, but now we are choosing a subgroup of three. That means n=6 and k=3:

solving a combinatorics problem

From here, you’ll notice that the 3*2*1 in the bottom cancels with the 6 in the top, leaving you with 5*4 = 20 in the numerator:

combination problem on the gmat answer

That leaves us with:

6C3 = 20 combinations of three cheeses

With 15 cases in the first situation and 20 in the second, the total is 35 cases, and our final answer is D. 

Next time, we’ll talk about what happens when we have permutations with repeat elements.

In the meantime, as an exercise, scroll back up and return to the 10-painting problem I presented earlier and see if you can find the answer. Bonus question: redo the problem with a subgroup of 6 paintings instead of 4 paintings. Try to anticipate: do you imagine we’ll have more combinations in this new case or fewer?

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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A Continuation of GMAT Permutation Math
Posted on
18
Feb 2021

A Continuation of Permutation Math

By: Rich Zwelling (Apex GMAT Instructor)
Date: 16 Feb 2021

Review of example from last post

Last time, when we started our discussion of GMAT Combinatorics, we gave a brief example of GMAT permutations in which we had five paintings and asked how many arrangements could be made on a wall with those paintings. As it turns out, no complicated combinatorics formula is necessary. You can create an easy graph with dashes and list five options for the first slot, leaving four for the second slot, and so on:

_5_  _4_ _3_ _2_ _1_

Then multiply 5*4*3*2*1 to get 120 arrangements of the five paintings. Remember you could see this notationally as 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)

More permutation math

But there could be fewer slots then items. Take the following combinatorics practice problem:

At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge?

A) 120
B) 240
C) 360
D) 480
E) 600

First, as a review, how do we know this is a PERMUTATION and not a COMBINATION? Because order matters. In the previous problem, the word “arrangements” gave away that we care about the order in which items appear. In this problem, we’re told that we’re interested in the “orderings” of cheeses. Cheddar followed by gouda would be considered distinct from gouda followed by cheddar. (Look for signal words like “arrangements” or “orderings” to indicate a PERMUTATION problem.)

In this case, we must consider the options of three or four cheeses separately, as they are independent (i.e. they cannot both happen). But for each case, the process is actually no different from what we discussed last time. We can simply consider each case separately and create dashes (slots) for each option. In the first case (three cheeses), there are six options for the first slot, five for the second, and four for the third:

_6_  _5_  _4_

We multiply those together to give us 6*5*4 = 120 possible ways to present three cheeses. We do likewise for the four-cheese case:

_6_  _5_  _4_  _3_

We multiply those together to give us 6*5*4*3 = 360 possible ways to present four cheeses.

Since these two situations (three cheeses and four cheeses) are independent, we simply add them up to get a final answer of 120+360 = 480 possible orderings of cheeses, and the correct answer is D. 

You might have also noticed that there’s a sneaky arithmetic shortcut. You’ll notice that you have to add 6*5*4 + 6*5*4*3. Instead of multiplying each case separately, you can factor out 6*5*4 from the sum, as follows:

6*5*4 + 6*5*4*3

= 6*5*4 ( 1 + 3)

= 6*5*4*4

= 30*16 OR 20*24

= 480

Develop the habit of looking for quick, efficient ways of doing basic arithmetic to bank time. It will pay off when you have to do more difficult questions in the latter part of the test. 

Now that we have been through GMAT permutations, next time, I’ll give this problem a little twist and show you how to make it a COMBINATION problem. Until then…

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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Featured Video Play Icon
Posted on
17
Feb 2021

Data Sufficiency: Area of a Triangle Problem

Hey guys! Today we’re checking out a geometry Data Sufficiency problem asking for the area of a triangle, and while the ask might seem straightforward, it’s very easy to get caught up in the introduced information on this problem. And this is a great example of a way that the GMAT can really dictate your thought processes via suggestion if you’re not really really clear on what it is you’re looking for on DS. So here we’re looking for area but area specifically is a discrete measurement; that is we’re going to need some sort of number to anchor this down: whether it’s the length of sides, or the area of a smaller piece, we need some value!

Begin with Statement #2

Jumping into the introduced information, if we look at number 2, because it seems simpler, we have x = 45 degrees. Now you might be jumping in and saying, well, if x = 45 and we got the 90 degree then we have 45, STOP. Because if you’re doing that you missed what I just said. We need a discrete anchor point. The number of degrees is both relative in the sense that the triangle could be really huge or really small, and doesn’t give us what we need. So immediately we want to say: number 2 is insufficient. Rather than dive in deeply and try and figure out how we can use it, let’s begin just by recognizing its insufficiency. Know that we can go deeper if we need to but not get ourselves worked up and not invest the time until it’s appropriate, until number 1 isn’t sufficient and we need to look at them together.

Consider Statement #1

Number 1 gives us this product BD x AC = 20. Well here, we’re given a discrete value, which is a step in the right direction. Now, what do we need for area? You might say we need a base and a height but that’s not entirely accurate. Our equation, area is 1/2 x base x height, means that we don’t need to know the base and the height individually but rather their product. The key to this problem is noticing in number 1 that they give us this B x H product of 20, which means if we want to plug it into our equation, 1/2 x 20 is 10. Area is 10. Number 1 alone is sufficient. Answer choice A.

Don’t Get Caught Up With “X”

If we don’t recognize this then we get caught up with taking a look at x and what that means and trying to slice and dice this, which is complicated to say the least. And I want you to observe that if we get ourselves worked up about x, then immediately when we look at this BD x AC product, our minds are already in the framework of how to incorporate these two together. Whereas, if we dismiss the x is insufficient and look at this solo, the BD times AC, then we’re much more likely to strike upon that identity. Ideally though, of course, before we jump into the introduced information, we want to say, well, the area of a triangle is 1/2 x base x height. So, if I have not B and H individually, although that’ll be useful, B x H is enough. And then it’s a question of just saying, well, one’s got what we need – check. One is sufficient. Two doesn’t have what we need – isn’t sufficient, and we’re there. So,

I hope this helped. Look for links to other geometry and fun DS problems below and I’ll see you guys soon. Read this article about Data sufficiency problems and triangles next to get more familiar with this type of GMAT question.

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Posted on
16
Feb 2021

Triangles With Other shapes

By: Rich Zwelling (Apex GMAT Instructor)
Date: 16 Feb 2021

As discussed before, now that we’ve talked about the basic triangles, we can start looking at how the GMAT can make problems difficult by embedding triangles in other figures, or vice versa. 

Here are just a few examples, which include triangles within and outside of squares, rectangles, and circles:

triangles in other shapes GMAT article

Today, we’ll talk about some crucial connections that are often made between triangles and other figures, starting with the 45-45-90 triangle, also known as the isosceles right triangle.

You’ve probably seen a rectangle split in two along one of its diagonals to produce two right triangles:

triangles in other shapes gmat article gmat probelm

But one of the oft-overlooked basic geometric truths is that when that rectangle is a square (and yes, remember a square is a type of rectangle), the diagonal splits the square into two isosceles right triangles. This makes sense when you think about it, because the diagonal bisects two 90-degree angles to give you two 45-degree angles:

triangles in other shapes gmat article, 45 45 90 degree angle

(For clarification, the diagonal of a rectangle is a bisector when the rectangle is a square, but it is not a bisector in any other case.)

Another very common combination of shapes in more difficult GMAT Geometry problems is triangles with circles. This can manifest itself in three common ways:

  1. Triangles created using the central angle of a circle

triangle in a circle, gmat geometry article

In this case, notice that two of the sides of the triangle are radii (remember, a radius is any line segment from the center of the circle to its circumference). What does that guarantee about the triangle?

Since two side are of equal length, the triangle is automatically isosceles. Remember that the two angles opposite those two sides are also of equal measure. So any triangle with the center of the circle as one vertex and points along the circumference as the other two vertices will automatically be an isosceles triangle.

2. Inscribed triangles

triangle inscribed in circle, gmat problem

An inscribed triangle is any triangle with a circle’s diameter as one of its sides and a vertex along the circumference. And a key thing to note: an inscribed triangle will ALWAYS be a right triangle. So even if you don’t see the right angle marked, you can rest assured the inscribed angle at that third vertex is 90 degrees.

3. Squares and rectangles inscribed in circles

rectangle in circle, gmat geometry

What’s important to note here is that the diagonal of the rectangle (or square) is equivalent to the diameter of the circle.

Now that we’ve seen a few common relationships between triangles and other figures, let’s take a look at an example Official Guide problem:

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A) 19,200
B) 19,600
C) 20,000
D) 20,400
E) 20,800

Explanation

The diagonal splits the rectangular park into two similar triangles:

triangle in other shapes gmat problem

Use SIGNALS to avoid algebra

It can be tempting to then jump straight into algebra. The formulas for perimeter and diagonal are P = 2L + 2W an D2 = L2 + W2, respectively, where L and W are the length and width of the rectangle. The second formula, you’ll notice, arises out of the Pythagorean Theorem, since we now have two right triangles. We are trying to find area, which is LW, so we could set out on a cumbersome algebraic journey.

However, let’s try to use some SIGNALS the problem gives us and our knowledge of how the GMAT operates to see if we can short-circuit this problem.

We know the GMAT is fond of both clean numerical solutions and common Pythagorean triples. The large numbers of 200 for the diagonal and 560 for the perimeter don’t change that we now have a very specific rectangle (and pair of triangles). Thus, we should suspect that one of our basic Pythagorean triples (3-4-5, 5-12-13, 7-24-25) is involved.

Could it be that our diagonal of 200 is the hypotenuse of a 3-4-5 triangle multiple? If so, the 200 would correspond to the 5, and the multiplying factor would be 40. That would also mean that the legs would be 3*40 and 4*40, or 120 and 160.

Does this check out? Well, we’re already told the perimeter is 560. Adding 160 and 120 gives us 280, which is one length and one width, or half the perimeter of the rectangle. We can then just double the 280 to get 560 and confirm that we do indeed have the correct numbers. The length and width of the park must be 120 and 160. No algebra necessary.

Now, to get the area, we just multiply 120 by 160 to get 19,200 and the final answer of A.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

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Combinatorics: Permutations and Combinations Intro
Posted on
11
Feb 2021

Combinatorics: Permutations and Combinations Intro

By: Rich Zwelling (Apex GMAT Instructor)
Date: 11 Feb 2021

GMAT Combinatorics. It’s a phrase that’s stricken fear in the hearts of many of my students. And it makes sense, because so few of us are taught anything about it growing up. But the good news is that, despite the scary title, what you need to know for GMAT combinatorics problems is actually not terribly complex.

To start, let’s look at one of the most commonly asked questions related to GMAT combinatorics, namely the difference between combinatorics and permutations

Does Order Matter?

It’s important to understand conceptually what makes permutations and combinations differ from one another. Quite simply, it’s whether we care about the order of the elements involved. Let’s look at these concrete examples to make things a little clearer:

Permutations example

Suppose we have five paintings to hang on a wall, and we want to know in how many different ways we can arrange the paintings. It’s the word “arrange” that often gives away that we care about the order in which the paintings appear. Let’s call the paintings A, B, C, D, and E:

ABCDE
ACDEB
BDCEA

Each of the above three is considered distinct in this problem, because the order, and thus the arrangement, changes. This is what defines this situation as a PERMUTATION problem. 

Mathematically, how would we answer this question? Well, quite simply, we would consider the number of options we have for each “slot” on the wall. We have five options at the start for the first slot:

_5_  ___ ___ ___ ___

After that painting is in place, there are four remaining that are available for the next slot:

_5_  _4_ ___ ___ ___

From there, the pattern continues until all slots are filled:

_5_  _4_ _3_ _2_ _1_

The final step is to simply multiply these numbers to get 5*4*3*2*1 = 120 arrangements of the five paintings. The quantity 5*4*3*2*1 is also often represented by the exclamation point notation 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!)

Combinations example

So, what about COMBINATIONS? Obviously if we care about order for permutations, that implies we do NOT care about order for combinations. But what does such a situation look like?

Suppose there’s a local food competition, and I’m told that a group of judges will taste 50 dishes at the competition. A first, a second, and a third prize will be given to the top three dishes, which will then have the honor of competing at the state competition in a few months. I want to know how many possible groups of three dishes out of the original 50 could potentially be selected by the judges to move on to the state competition.

The math here is a little more complicated without a combinatorics formula, but we’re just going to focus on the conceptual element for the moment. How do we know this is a COMBINATION situation instead of a permutation question? 

It’s a little tricky, because at first glance, you might consider the first, second, and third prizes and believe that order matters. Suppose that Dish A wins first prize, Dish B wins second prize, and Dish C wins third prize. Call that ABC. Isn’t that a distinct situation from BAC? Or CAB? 

Well, that’s where you have to pay very close attention to exactly what the question asks. If we were asking about distinct arrangements of prize winnings, then yes, this would be a permutation question, and we would have to consider ABC apart from BAC apart from CAB, etc. 

However, what does the question ask about specifically? It asks about which dishes advance to the state competition? Also notice that the question specifically uses the word “group,” which is often a huge signal for combinations questions. This implies that the total is more important than the individual parts. If we take ABC and switch it to BAC or BCA or ACB, do we end up with a different group of three dishes that advances to the state competition? No. It’s the same COMBINATION of dishes. 

Quantitative connection

It’s interesting to note that there will always be fewer combinations than permutations, given a common set of elements. Why? Let’s use the above simple scenario of three elements as an illustration and write out all the possible permutations of ABC. It’s straightforward enough to brute-force this by including two each starting with A, two each starting with B, etc:

ABC
ACB
BAC
BCA
CAB
CBA

But you could also see that there are 3*2*1 = 3! = 6 permutations by using the same method we used for the painting example above. Now, how many combinations does this constitute? Notice they all consist of the same group of three letters, and thus this is actually just one combination. We had to divide the original 6 permutations by 3! to get the correct number of permutations.

Next time, we’ll continue our discussion of permutation math and begin a discussion of the mechanics of combination math. 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements

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triangle inequality rule
Posted on
09
Feb 2021

Triangle Inequality Rule

By: Rich Zwelling (Apex GMAT Instructor)
Date: 9 Feb 2021

One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.

Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:

Suppose we have a triangle that has two sides of length 3 and 5:

triangles inequalities 1

What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.

So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]

If you’d like to see that put into words:

**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**

It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)

Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.

triangle inequalities 2

 

If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:

triangle inequalities 3

The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)

Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:

If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five

Strategy: Eliminate Answers

As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).

Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.

Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.

Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.

Note:

It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

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gmat in Atlanta
Posted on
04
Feb 2021

Taking the GMAT Exam in Atlanta

Table of Contents:

  1. Who administers the GMAT test?
  2. What does the test center look like?
  3. Where are the test centers located?
  4. Test center holidays
  5. Top MBA programs in the area
  6. Tips
  7. Test Day FAQs

About ¾ of the way through your extensive GMAT prep you should begin to start planning your test day, including scheduling the test, preparing your trip to the test center, and even pre-visiting the test center so that you know exactly where it is. This guide is here to offer you all the required information related to taking the GMAT in Atlanta.

Who administers the GMAT test in Atlanta?

Pearson Professional Centers – administers the GMAT and EA exam on behalf of the GMAC. To find out more about the Pearson Professional test centers visit https://www.pearson.com/us/.

What does the test center look like?

A Pearson Professional Center will include individual testing areas for each test taker with a separation screen between each test-taker. 

Where are the test centers in Atlanta?

These are the 2 top locations where test-takers generally had the best experience:

Pearson Professional Centers-Atlanta GA

2 Corporate Blvd NE
Suite 150
Atlanta, Georgia 30329
United States

 

By car:gmat in atlanta 1

From Downtown Atlanta (13 minutes):

  • Get on I-75 N/I-85 N from Ellis St NE
  • Continue on I-75 N/I-85 N to North Druid Hills. Take exit 89 from I-85 N
  • Continue on I- 85 Access Rd/I – 85 Frontage Rd to your destination in Brookhaven
Test-takers’ review:

This test center was rated 3.1 by Google reviewers. They mentioned that the building was easy to find and the staff was exceptionally polite and professional. They generally had a good test-taking experience.

Pearson Professional Centers-Atlanta (North) GA

1117 Perimeter Center West
Suite 311 East
Atlanta, Georgia 30338
United States

By car:gmat in atlanta 2

 

From Downtown Atlanta (18 minutes):

  • Get on I-75 N/I-85 N from Ellis St NE
  • Continue on I-75 N/I-85 N. Take GA-400 N to Abernathy Rd NE in Sandy Springs. Take exit 5 from US-19 N
  • Continue on Abernathy Rd NE to your destination
Test-takers’ review:

This test center was rated 3.1 by Google reviewers. They mentioned that the staff was efficient and the checking-in process was a breeze. The office was also clean, neat, and organized.

Test center holidays

 

The most popular times for GMAT preparation and test-taking are during the holiday seasons. Be mindful of dates that you will not be able to take the GMAT or EA at any of the test centers mentioned above. Pearson test centers are closed during the following dates:

  • 1 Jan – New Year’s Day 
  • 2 Apr – Good Friday   
  • 5 Apr – Easter Monday 
  • 3 May – May Day 
  • 31 May – Late May Bank Holiday   
  • 30 Aug – August Bank Holiday
  • 25 Dec – Christmas Day
  • 26 Dec – Boxing Day 
  • 27 Dec – Christmas Holiday
  • 28 Dec – Boxing Day Holiday

Top MBA programs in Atlanta:

  1. Scheller College of Business- Georgia Institute of Technology

Tips:

  • During the test there will not be complete silence – you will be able to hear noise from other test takers so it is best to prepare for this by studying for the exam in similar scenarios. This can prepare you for any distractions (such as coughing, sneezing, or computer clicking sounds) that might occur while taking your exam.  
  • Try to spend some time actually prepping in the lobby of the test center weeks/days in advance of your exam date. Since the place will be familiar to you come test day this can help curb test anxiety should you have any.

Test Day FAQs

Here are the top 5 questions that clients ask us about exam day information:

  • Are you allowed to listen to music while taking the GMAT exam?

You are not allowed to listen to music while taking the GMAT exam and you are not allowed to wear earphones as well. 

 

  • What should I do if I fall sick on the exam day?

If you do not feel well come exam day you will have to make the decision as to whether or not you can take the test and perform at your best. Most people will not be able to do this so it will be best to cancel it. If you do so on the day of the exam you will incur a loss of your full $250 exam fee. If you cancel the exam 7 days in advance you will be charged a penalty of $50. If it is the first time that you will sit the exam and you are up for sitting through a 4 hour test, this may be a good opportunity to experience the test as you have the ability to cancel the score right afterwards if you are unhappy with it. Ultimately, it is best to take the GMAT when you are feeling your best as this will result in your optimum test performance. 

 

  • What can I bring with me to the test center?

You are allowed in the test center with the following:

  • GMAT approved identification
  • Appointment confirmation letter or email you received from Pearson VUE
  • Prescription eyeglasses
  • Light sweater or light non-outerwear jacket
  • Comfort items only if they were pre-approved as an accommodation received in advance

Any additional personal belongings that you bring with you such as your cell phone, bag, snacks, and earphones will need to be stored in one of the provided lockers. You may eat your snack during the breaks. Any cell phone use throughout the test time (including breaks) is prohibited. 

The test center will provide you with everything that you need in order to take the test including scratch paper and a pencil. 

 

  • Should I wear a mask during the exam?

At the test centers above they strongly recommend that you wear a face mask or some type of face-covering in the test center and for the duration of your test to protect yourself and others. Test centers do not provide face masks for candidates. 

Please note that if you have any flu-like symptoms upon arrival at the test center, you may be requested to reschedule your exam for another time when you are in full health.

  • What can I expect at the test center?

A usual test center is typically quite small. Once you arrive you will have to provide the administrator with the relevant documents and while these are being processed you will be asked to wait in the waiting area. In this area, you can still access all your personal belongings up until you are called into the testing room. 

Once in the room, you will be allocated an individual exam station where you will find a computer. 

Find more FAQs: HERE

 

For any questions or comments please reach out to us at www.apexgmat.com.

To speak to an Apex instructor about your GMAT prep, schedule a call HERE.

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