The basics of GMAT Combinatorics
Posted on
24
Jun 2021

The Basics of GMAT Combinatorics

By: Apex GMAT
Contributor: Svetozara Saykova
Date: 24th June 2021

Combinatorics can seem like one of the most difficult types of questions to come across on the GMAT. Luckily there are not many of them within the exam. Still these questions make up the top level of scoring on the test and therefore it is best if you are well equipped to solve them successfully, especially if you are aiming for a 700+ score. The most important rule to follow when considering this question type is the “Fundamental Counting Principle” also known as the “Counting Rule.” This rule is used to calculate the total number of outcomes given by a probability problem. 

The most basic rule in Combinatorics is “The Fundamental Counting Principle”. It states that for any given situation the number of overall outcomes is equal to the product of the number of each discrete outcome.

Let’s say you have 4 dresses and 3 pairs of shoes, this would mean that you have 3 x 4 = 12 outfits. The Fundamental Counting Principle also applies for more than 2 options. For example, you are at the ice cream shop and you have a variety of 5 flavors, 3 types of cones and 4 choices for toppings. That means you have 5 x 3 x 4 = 60 different combinations of single-scoop ice creams. 

The Fundamental Counting Principle applies only for choices that are independent of one another. Meaning that any option can be paired with any other option and there are no exceptions. Going back to the example, there is no policy against putting sprinkles on strawberry vanilla ice cream because it is superb on its own. If there were, that would mean that this basic principle of Combinatorics would not apply because the combinations (outcomes) are dependent. You could still resort to a reasoning solution path or even a graphical solution path since the numbers are not so high. 

Let’s Level Up a Notch

The next topic in Combinatorics is essential to a proper GMAT prep is  permutations. A permutation is a possible order in which you put a set of objects.

Permutations

There are two subtypes of permutations and they are determined by whether repetition is allowed or not.

  • Permutations with repetition allowed

When there are n options and r number of slots to fill, we have n x n x …. (r times) = nr permutations. In other words, there are n possibilities for the first slot, n possibilities for the second and so on and so forth up until n possibilities for position number r.

The essential mathematical knowledge for these types of questions is that of exponents

To exemplify this let’s take your high school locker. You probably had to memorize a 3 digit combination in order to unlock it. So you have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) for 3 available slots. The total number of locker passwords you can have is 103 = 1,000. 

  • Permutation without repetition allowed 

When repetition is restricted in the given GMAT problem, we would have to reduce the number of available choices for each position. 

Let’s take the previous example and add a restriction to the password options – you cannot have repeating numbers in your locker password. Following the “we reduce the options available each time we move to the next slot” rule, we get 10x9x8 = 720 options for a locker combination (or mathematically speaking permutation). 

To be more mathematically precise and derive a formula we use the factorial function (n!). In our case we will take all the possible options 10! for if we had 10 positions available  and divide them by 7!, which are the slots we do not have. 

10! =  10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 

7! =  7 x 6 x 5 x 4 x 3 x 2 x 1 

And when we divide them (7 x 6 x 5 x 4 x 3 x 2 x 1) cancels and we are left with 10 x 9 x 8 = 720. 

Pro tip: Taking problems and deeply examining them by running different scenarios, and changing some of the conditions or numbers is a great way to train for the GMAT. This technique will allow you to not only deeply understand the problem but also the idea behind it, and make you alert for what language and piece of information stands for which particular concept. 

So those are the fundamentals, folks. Learning to recognize whether order matters and whether repetition is allowed is essential when it comes to Combinatorics on the GMAT. Another vital point is that if you end up with an endless equation which confuses you more than helps, remember doing math on the GMAT Quant section is not the most efficient tactic. In fact, most of the time visualizing the data by putting it into a graph or running a scenario following your reasoning are far more efficient solution paths. 

Feeling confident and want to test you GMAT Combinatorics skills? Check out this GMAT problem and try solving it. Let us know how it goes!

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When Probability Meets Combinatorics: One Problem, Two Approaches article
Posted on
16
Mar 2021

When Probability Meets Combinatorics: One Problem, Two Approaches

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th March, 2021

Now, we’d like to take a look at an Official GMAT Probability problem to pull everything together. The following is a good example for two reasons:

 1. It illustrates a quirky case that is difficult more conceptually than mathematically, and thus is better for the GMAT.

 2. It can be tackled either through straight probability or through a combination of probability and combinatorics.

Here’s the question:

Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

A) 1/24
B)
1/8
C) 1/4
D) 1/3
E) 3/8

First, as always, give the problem a shot before reading on for the explanation. If possible, see if you can tackle it with both methods (pure probability and probability w/ combinatorics). 

Explanation #1:

First, we’ll tackle pure probability. Let’s label the letters A, B, C, and D, and let’s say that A is the letter we wish to match with its correct envelope. The other three will be matched with incorrect envelopes. We now must examine the individual probabilities of the following events happening (green for correct, red for incorrect):

_A_   _B_   _C_   _D_

For the above, each slot represents a letter matched with an envelope. There are four envelopes and only one is correct for letter A. That means Tanya has a 1/4 chance of placing letter A in its correct envelope:

_1/4__   _B_   _C_   _D_

We now desire letter B to be placed in an incorrect envelope. Two of the remaining three envelopes display incorrect addresses, so there is a 2/3 chance of that happening:

_1/4__   _2/3_   _C_   _D_

We then desire letter C to also be placed in an incorrect envelope. Only one of the remaining two envelopes displays an incorrect address, so there is a 1/2 chance of that happening:

_1/4__   _2/3_   _1/2_   _D_

At that point, the only remaining option is to place the last remaining letter in the last remaining envelope (i.e. a 100% chance, so we place a 1 in the final slot):

_1/4__   _2/3_   _1/2_   _1_

Multiplying the fractions, we can hopefully see that some cancelling will occur:

¼ x ⅔ x ½ x 1

= 1 x 2 x 1
   ———–
   4 x 3 x 2

= 1/12

But lo and behold, 1/12 is not in our answer choices. Did you figure out why?

We can’t treat letter A as the only possible correct letter. Any of the four letters could possibly be the correct one. However, the good news is that in any of the four cases, the math will be exactly the same. So all we have to do is take the original 1/12 we just calculated and multiply it by 4 to get the final answer: 4 x 1/12 = 4/12 = 1/3. The correct answer is D.

Explanation #2:

So what about a combinatorics approach?

As we’ve discussed in our previous GMAT probability posts, all probability can be boiled down to Desired Outcomes / Total Possible Outcomes. And as we discussed in our posts on GMAT combinatorics, we can use factorials to figure out the total possible outcomes in a situation such as this, which is actually a simple PERMUTATION. There are four envelopes, so for the denominator of our fraction (total possible outcomes), we can create a slot for each envelope and place a number representing the letters in each slot to get:

_4_  _3_  _2_  _1_  =  4! = 24  possible outcomes

This lets us know that if we were to put the four letters into the four envelopes at random, as the problem says, there would be 24 permutations, giving us the denominator of our fraction (total possible outcomes). 

So what about the desired outcomes? How many of those 24 involve exactly one correctly placed letter? Well, let’s again treat letter A as the correctly placed letter. Once it’s placed, there are three slots (envelopes) left: 

___  ___  ___ 

But the catch is: the next envelope has only two letters that could go into it. Remember, one of the letters correctly matches the envelope in address, and we want a mismatch:

_2_  ___  ___ 

Likewise, that would leave two letters available for the next envelope, but only one of them would have the wrong address:

_2_  _1_  ___ 

And finally, there would be only one choice left for the final envelope:

_2_  _1_  _1_ 

That would mean for the correctly-placed A letter, there are only two permutations in which each of the other letters is placed incorrectly:

_2_ x  _1_ x  _1_ = 2 possible outcomes.

But as before, we must consider that any of the four letters could be the correct letter, not just letter A. So we must multiply the 2 possible outcomes by four to get 8 desired outcomes involving exactly one letter being placed in its correct envelope. That gives us our numerator of Desired Outcomes. Our denominator, remember, was 24 total possible outcomes. So our final answer, once again, is 8/24 = 1/3.

This is a great example of how GMAT combinatorics can intersect with probability.

To tide you over until next time, give this Official GMAT problem a try. It will also give a nice segue into Number Theory, which we’ll begin to talk more about going forward. Explanation next time…

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

 

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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Posted on
18
Feb 2021

A Continuation of Permutation Math

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
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

 

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 2021

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Combinatorics & Permutations
Posted on
11
Feb 2021

Introduction to Combinatorics & Permutations

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. 

 

Contributor: Rich Zwelling, Apex GMAT Instructor

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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