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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An “Undesired” Approach to GMAT Probability gmat article
Posted on
11
Mar 2021

An “Undesired” Approach to GMAT Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th March, 2021

In our last post, we discussed a solution for the following question, which is a twist on an Official Guide GMAT probability problem:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

We also mentioned that there was an alternate way to solve it. Did you find it? In truth, it relates to something we discussed in a previous post we did on GMAT Combinatorics, specifically Combinations with Restrictions. In that post, we discussed the idea of considering combinations in which you’re not interested. It might seem counterintuitive, but if you subtract those out from the total number of combinations possible, you’re left with the number of combinations in which you are interested:

You can actually do something similar with probability. Take the following basic example:

Suppose I told you to flip a fair coin five times, “fair” meaning that it has an equal chance of landing heads-up or tails-up. I then wanted to know the probability that I flip at least one head. Now, when you think about it, the language “at least one” involves so many desired possibilities here. It could be 1 head, 2 heads, …, all the way up to 5 heads. I’d have to calculate each of those probabilities individually and add them up.

Or…

I could consider what is not desired, since the possibilities are so much fewer:

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

All of the above must add to 100% or 1, meaning all possible outcomes. So why not figure out the probability that I get 0 heads (or all tails), and then subtract it from 100% or 1 (depending on whether I’m using a percentage or decimal/fraction)? I’ll then be left with all the possibilities in which I’m actually interested, without the need to do any more calculations.

Each time I flip the coin, there is a ½ chance that I flip a tail. This is the same each of the five times I flip the coin. I then multiply all of the probabilities together:

½ x ½ x ½ x ½ x ½ = 1 / 25  = 1 / 32

Another way to view this is through combinatorics. Remember, probability is always Desired outcomes / Total possible outcomes. If we start with the denominator, there are two outcomes each time we flip the coin. That means for five flips, we have 25 or 32 possible outcomes, as illustrated here with our slot method:

_2_  _2_  _2_  _2_  _2_ = 32

Out of those 32 outcomes, how many involve our (not) desired outcome of all tails? Well, there’s only one possible way to do that: 

_T_  _T_  _T_  _T_  _T_    ← Only 1 outcome possible

It really is that straightforward: with one outcome possible out of 32 total, the probability is 1/32 that we flip all tails. 

Now remember, that is our, not desired. Our desired is the probability of getting at least one head

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

So, since the probability of getting 0 heads (all tails) is 1/32, we simply need to subtract that from 1 (or 32/32) to get our final result. The probability that we flip at least one head if we flip a fair coin five times is 31/32.

Application to problem from previous post

So now, how do we work that into the problem we did last time? Well, in the previous post, we took a more straightforward approach in which we considered the outcomes we desired. But can we use the above example and consider not desired instead? Think about it and give it a shot before reading the explanation:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B) 5/8
C) 3/8
D) 5/64
E) 3/64

Explanation

In this question, our desired case is that Zelda does not attempt the problem. That means, quite simply, that our not desired case is that Zelda does get to attempt it. This requires us analytically to consider how such a case would arise. Let’s map out the possibilities with probabilities:

An “Undesired” Approach to GMAT Probability treeNotice that two complementary probabilities are presented for each box. For example, since there is a 1/4 chance Xavier solves the problem (left arrow), we include the 3/4 probability that he does not solve the problem (right arrow). 

If Zelda does get to attempt it, it’s clear from the above that first Xavier and Yvonne must each not solve it. There is a 3/4 and a 1/2 chance, respectively, of that happening. This is also a dependent situation. Xavier must not solve AND Yvonne must not solve. Therefore, we will multiply the two probabilities together to get ¾ x ½ = ⅜. So there is a 3/8 chance of getting our not desired outcome of Zelda attempting the problem.

So, we can finally subtract this number from 1 (or 8/8) and see that there is a 5/8 chance of Zelda not getting to attempt the problem. The correct answer is B.

Next time, we’ll discuss how GMAT Probability and Combinatorics can combine to form some interesting problems…

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

Read more
GMAT Combinations with Restrictions Article
Posted on
04
Mar 2021

Combinations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 4th March, 2021

In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS.

Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same.

Let’s start with a basic example. Suppose I were to give you the following problem:

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible?

This is a straightforward combination problem. (And we know it’s a COMBINATION situation because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.)

We would simply use the combination math discussed in our Intro to Combination Math post:

                         10!
 10C3 =       ————-
                     3! (10-3!)

 

   10!
———
3! (7!)

 

10*9*8
———
3!

 

10*9*8
———
3*2*1

= 120 Combinations 

However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…):

The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman?

A) 60
B) 75
C) 85
D) 90
E) 95

In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work. 

But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman). 

We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3  or 120 combinations possible. 

But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that?

Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men:

                       7!
7C3 =       ————-
                 3! (7-3!)

 

  7!
———
3! (4!)

 

7*6*5
———
    3!

= 7*5 = 35 Combinations involving all men

So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C. 

Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions.

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

Read more
Permutations with Restrictions GMAT Article
Posted on
02
Mar 2021

Permutations with Restrictions

By: Rich Zwelling, Apex GMAT Instructor
Date: 2nd 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
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

Read more
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: 25th 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
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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An Intro to Combination Math GMAT Article
Posted on
23
Feb 2021

An Intro to Combination Math

By: Rich Zwelling, Apex GMAT Instructor
Date: 23rd February, 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
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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A continuation of permutation math on the GMAT
Posted on
18
Feb 2021

A Continuation of Permutation Math

By: Rich Zwelling, Apex GMAT Instructor
Date: 16th February, 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
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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Combinatorics: Permutations and Combinations Intro On the GMAT
Posted on
11
Feb 2021

Combinatorics: Permutations and Combinations Intro

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th February, 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
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

Read more
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Posted on
16
Apr 2020

Combinatorics GMAT Problem: Movie Night

Today we’ve got a fairly straightforward GMAT combinatorics problem. If you’ve been self-prepping in a rigorous, let me review the rules sort of way, you’ll pick up that there’s orders, combinations here and you might be inclined to really dive in. What’s my combinations formula? What’s my permutation formula? How do I know which is which? Then plug in numbers.

While that will get you there understand that most GMAT combinatorics problems are more about being familiar with combinatorics than any really heavy duty math. That is because the number of people who are taking the GMAT are generally more familiar with Algebra or Geometry.

Combinatorics & The GMAT

Combinatorics, by virtue of being less known, is considered more valuable. It is scored more highly than problems of similar complexity in Algebra or Geometry. So you’re really being rewarded just for knowing basic combinatorics and in fact most permutation/combination problems fall into this basic category. The good news here is that you can use your reasoning to solve this problem without being burdened by the formal combinatorics formulas.

Solving The Problem

Let’s take a look at this problem. John’s having a movie night. We need to ask ourselves a series of pivot questions. How many different movies can John show first?

Well there’s 12 movies, he could show any of the 12. Leaving 11 movies to be shown second, any of 11. 10, 9. So the answer is 12x11x10x9 or 11,880. But even this math is a lot to do. Notice that by walking it through as a story, as a narrative, we don’t need to cancel out the 7 6 5 4 3 2 1. We don’t need to worry about division or anything else. We just know that there’s 4 movies and each time, each step we take, there’s one less movie available. Here we have this product 12 times 11 times 10 times 9, but we don’t really want to be forced to process this and so we can look for features that allow us to skip doing that heavy math.

Transforming The Numbers

We’ve got this really neat triangular shape in the answer choices where each answer has a different number of digits in it. 12, 11, 10, 9, we can look at and say on average each one’s about 10. The 9 and the 12 sort of compensate, but overall we’re going to have something that’s close to 10 times 10 times 10 times 10.

That is our answer should be somewhere around 10,000 or possibly a little more because we have an 11 and a 12 offset only by a 9. So what we’re looking for is something in that just above 10,000 range this prevents us from doing the math and very rapidly lets us look at those four movies, those numbers 12 11 10 9 and zero in on that 11 880 number.

Problem Form

Try it again with a similar number. Notice that you can’t do this with a hundred different movies selecting 17 of them. The math, the numbers would be too cumbersome.

The GMAT is really restricted here and you should restrict yourself to ones that are reasonable to keep processed in your head without doing heavy duty math. Similarly, notice how this one clusters around ten, it doesn’t have to cluster around ten, but when you’re rewriting this problem think about that clustering and think about how your knowledge of common powers or how other identities can help you rapidly get to an answer because the GMAT will present you with numbers that have a neat clean way to jump from your understanding directly to the answer without all that messy math in between.

This is Mike for Apex GMAT with your problem of the day.

If you enjoyed this combinatorics GMAT problem, try more GMAT practice problems:  Remainder Number Theory 

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