GMAT Combinatorics. This phrase has stricken fear in the hearts of many GMAT test takers. You never know when a challenging combination or permutation question will pop up three-quarters of the way through your GMAT exam to wreak havoc on your score.

Combinatorics is the branch of mathematics studying the enumeration, combination, and permutation of sets of elements and the mathematical relations that characterize their properties.

It is not as commonly tested as algebra fundamentals or number properties, which is why many GMAT test takers perceive it as challenging. The good news is that, despite the scary title, what you need to know for GMAT combinatorics problems is actually not terribly complex.

## The Basics of GMAT Combinatorics

Every combinatorics question on the GMAT involves one or more of** three fundamental combinatorics tasks: ordering, permutations, and combinations.**

### The Ordering Task

The *ordering* task is “lining up” elements spatially or temporally (in time).

**Spatial**: How many ways can 7 cars be lined up in 7 parking spaces?**Temporal**: In how many different orders can 7 contestants perform in a talent show?

There is no mathematical difference between these two scenarios. Both can be represented with the same standard formula.

### The Permutations Task: Selecting and Ordering

The *permutations* task is “lining up” a *subgroup* of elements spatially or temporally. We may refer to permutations as “selecting and ordering.” We will order elements in time or space, but only the *subgroup* of elements that has been selected out of the larger group. As with the ordering task, spatial and temporal instances of the permutations task are mathematically identical.

**Spatial**: How many ways can 4 cars from a collection of 7 cars be lined up in 7 parking spaces?**Temporal**: The judges of a talent show will create a performance schedule for 4 finalists selected from among 7 semifinalists. How many different schedules can be created?

### The Combinations Task: Selecting Without Ordering

The *combinations* task is selecting a subgroup of elements out of a larger group, without regard for their order. We may refer to combinations as “selecting without ordering,” or simply “selecting.” In combinations, the spatial/temporal distinction disappears altogether. Since nothing is being “lined up,” there is no time or space where this “lining up” occurs. Selecting a subgroup out of a larger group is not a spatial or temporal operation.

**Spatial**: How many ways can 4 cars be selected from a collection of 7 cars?**Temporal**: How many ways can 4 finalists be chosen from among 7 semifinalists in a talent show?

## Difference Between Combinations and Permutations: Does Order Matter?

Half the challenge in combination and permutation questions is identifying whether it is a combination or permutation problem.

Quite simply, what makes permutations and combinations differ from one another is whether we care about the order of the elements involved.

It makes sense to think of permutations as “selecting, then ordering.” Permutations entails performing the combinations task and then performing the ordering task on the selected subgroup.

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: Fewer Combinations Than Permutations

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.

## How to Recognize and Solve GMAT Ordering Tasks

### What is an Ordering Task in Combinatorics?

Ordering tasks are a common type of question on the GMAT. In an ordering task, you are asked to determine the number of ways to order a set of objects. The number of ways to order n different objects is n!, or n factorial.

The best way to conceptualize the ordering task is as *filling slots*. Here’s an example question with a simple graphical representation:

How many ways can 7 cars be lined up in 7 parking spaces?

___ ___ ___ ___ ___ ___ ___

We can imagine these seven slots as the seven parking spaces. How many different options do we have for filling the first slot? Well, since we haven’t assigned any cars to any slots yet, we have all seven cars available.

___ ___ ___ ___ ___ ___ ___

7

What next? No matter which of the seven cars was assigned to the first slot, we have six cars left.

___ ___ ___ ___ ___ ___ ___

7 6

By now you might see where this is going. If we keep filling in the possible options for each slot, from left to right, we get this:

___ ___ ___ ___ ___ ___ ___

7 6 5 4 3 2 1

### The n! Formula for Ordering Tasks in Combinatorics

By the time we get to the last slot, we have assigned every car but the last one, so we have only one choice. Those who know their math notations may recognize the “descending integers” pattern formed in this scenario:

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1

This is just one example of the standard formula for the ordering task. The number of ways to order *n* different things (spatially or temporally) is *n*! (pronounced “n factorial”), the product of *n* multiplied by each positive integer less than *n*. If we had 10 cars instead of 7 cars, there would be 10! ways to line up those 10 cars in 10 parking spaces. If we had 16 cars, there would be 16! ways to line up those 16 cars in 16 parking spaces.

### Ordering Tasks with Identical Elements

Simple enough, but the ordering task can involve an extra twist: what if some of the elements being ordered are considered *identical* or *indistinguishable*? In other words, we count all orders in which these identical elements trade places with each other (while other elements stay put) as identical orders.

Consider the word INTUITION. How many different ways can the letters in this word be arranged? You might be quick to answer “9 factorial,” since there are 9 letters. However, this would overcount the number of *different* ways the letters can be arranged, since any arrangements where identical letters trade places are not recognizably different.

Let’s say we arrange the letters like this:

INTIOUNTI

Technically, those three I’s can swap places amongst themselves, and we would never be able to tell. Likewise, the two T’s or the two N’s can trade places with each other, and we would never be able to tell. This is true for *any* arrangement of the letters – not just in our example. However, the “n factorial” method counts each one of these identical arrangements separately, significantly overcounting the number of *different* arrangements.

The good news is that we don’t have to start from scratch: we can correct this overcount using our existing understanding of ordering. Let’s consider each group of identical letters as its own isolated group. Since there are three I’s, the number of ways to order the I’s independently is 3!

So for any given arrangement of the letters in the word INTUITION, there are 3! ways that the I’s can swap places amongst themselves without us noticing. Likewise, and simultaneously, there are 2! ways for the N’s to trade places and 2! ways for the T’s to trade places.

The “n factorial” counting method for ordering will count each of these indistinguishable arrangements separately. So we must divide the “n factorial” number by the number of indistinguishable arrangements hiding within each *different* arrangement. Since the word INTUITION has 9 letters, 3 I’s, 2 N’s, and 2 T’s, the number of *different* arrangements for the letters is 9! / (3!*2!*2!).

When identical or indistinguishable elements are present, we divide the factorial of the full number of elements (n factorial) by the product of the factorials of each group of identical elements.

## How to Recognize and Solve GMAT Permutations Problems

To summarize where we are: Understanding combinatorics on the GMAT involves recognizing the type of problem you’re dealing with and applying the appropriate principles: ordering, permutations, or combinations. Let’s take a closer look at how to approach permutations problems.

In a permutations question, you are asked to find the number of ways to order a group of elements. The order of the elements matters in permutations questions, so the number of possible permutations can be quite large.

The permutations task is “lining up” a subgroup of elements spatially or temporally.

We learned that it makes sense to think of performing the permutations task as performing the combinations and ordering tasks successively: counting ways to select a subgroup and then counting ways to order a selected subgroup. However, this understanding does not link up with the standard formula for permutations – at least not at first glance.

### Permutations as Ordering with a Stopping Point

To understand the standard formula for permutations, it’s better to think of permutations as “ordering with a stopping point.” Consider the following example:

Example: Lining Up Cars

How many ways can 4 cars from a collection of 7 cars be lined up in 4 parking spaces?

Let’s represent this scenario using the slot system from before:

___ ___ ___ ___

Since we are filling only four parking spaces, we will only use four slots. However, we still have seven options for the first slot!

___ ___ ___ ___

7

And we still have six options for the second slot, and so on:

___ ___ ___ ___

7 6 5 4

Once you’ve assigned these four cars you were asked to assign, you’re done. The number of ways to line up 4 cars from a collection of 7 cars in 4 parking spaces is 7 * 6 * 5 * 4 = 840

Permutations is like ordering with a stopping point, because we simply stop after assigning however many elements we were asked to assign.

### The Standard Formula for Permutations

Now let’s talk about the standard formula for permutations. The full number of elements from which you select your subgroup is always represented as n. In this example with the cars, n = 7. But depending on where in the world you learn math, the number of elements in the subgroup may be represented as r or k. We’ll stick with r in these articles.

Here’s the standard permutations formula for ordering r things from a group of n things:

**n!(n-r)!**

Since n is the number of things in the whole group, and r is the number of things in the selected group, (n – r) is the number of things not being selected. We begin with the n! from the ordering task, and then we divide this number by the number of ways to arrange the things not being selected: (n – r)!

### Connecting the Permutations Formula to the Slot-Filling Method

Let’s go back to the car example. We were asked to line up 4 cars from a collection of 7 cars, so n = 7 and r = 4. In this case, (n – r)! = (7 – 4)! = 3! Here’s how this connects to our “slot-filling” method:

___ ___ ___ ___ ___ ___ ___

7 6 5 4 3 2 1

The numbers after our stopping point correspond to the (n – r)! we divided out from the initial value of n! So there you have it: the permutations task is ordering with a stopping point.

There are a few more variants of permutations questions that you can see on the GMAT. Read more about permutations with repeat elements, and how to solve permutation problems with restrictions.

## How to Recognize and Solve GMAT Combination Problems

Conceptually, the combinations task is easy to understand: it’s about the number of ways to select a subgroup from a whole group.

### The Standard Formula for Combinations

As in the permutations task, the whole group is notated as *n*, and the subgroup is notated as *r*. The standard formula for combinations adds just one piece to the standard formula for permutations: the expression *r*! in the denominator:

Here is the Combinations standard formula:

**n!/r!(n-r)!**

Considering that *r* is the number of elements in the selected subgroup, and going back to our understanding of ordering, *r*! is the number of ways to order the elements in the selected subgroup.

Since the permutations task is “selecting and ordering” and combinations task is “selecting *without* ordering,” it makes sense that we get from permutations to combinations by *dividing out* the number of ways to order the things we selected.

Thus, the combinations standard formula counts the number of different *subgroups* of *r* things that can be selected from a whole group of *n* things.

While the permutations standard formula counts the number of different *orders* of *r* things selected from a whole group of *n* things.

The expression *r*! is the number of *orders per subgroup*. If we divide the number of *orders* (given by the permutations formula) by the number of *orders per subgroup* (*r*!), we obtain the number of *subgroups* (the combinations formula).

*number of orders/number of orders per subgroup = number or subgroups*

Let’s return one last time to our example with the cars:

How many ways can 4 cars be selected from a collection of 7 cars?

*n* = 7 and *r* = 4, so we can plug these values into the standard formula:

**n!/r!(n-r)!**

**7!/4!(7-4)! = 35**

In effect, we just performed the permutations task (counted the number of orders for four cars selected from seven cars) and divided out the number of orders per subgroup (4!) to obtain an answer of 35 possible subgroups of four cars.

Let’s put this connection between combinations and permutations in real terms. When we selected and ordered four cars from the group of seven (performing the permutations task), we might have done it like this:

Aston Martin Bentley Corvette DeLorean

But we also might have done it like this:

Corvette Aston Martin DeLorean Bentle

Same four cars, in different order. With our understanding of ordering, we know that there are 4! ways to line up these same four cars. Since the combinations task is about identifying *unique subgroups* rather than *unique orders*, we must divide out the number of ways to line up each different subgroup of *r* things: *r*!

If you’re on a roll, you might also want to take a look at how to solve combination problems with restrictions on the GMAT with example problems and pitfalls to avoid.

## How to Prep for GMAT Combinatorics

If you’ve powered through this extensive guide on GMAT Combinatorics, you already have the essential formulas for these questions. More importantly, I hope, you also have a deeper understanding of the relationships between these formulas.

However, to ace your GMAT exam questions on permutations and combinations, the formulas and the default solving mechanisms are simply not going to be enough. You also need to develop the habit of looking for quick, efficient ways of doing basic arithmetic to save time. It will pay off when you have to do more difficult questions in the latter part of the test.

If you find yourself scratching stuff on paper and doing extensive calculations, that is a signal that you’ve missed the bigger picture of the problem. Every problem on the GMAT exam, quant and otherwise, has multiple solution paths and some are more efficient than others. The moment you’re doing a lot of processing, that’s an indication that you’re not seeing the most efficient solution path.

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