# GRE Combinations and Permutations: Worked Examples | topin

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## Define one outcome before choosing a formula

[ETS](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html) lists combinations and permutations within GRE Quant’s counting methods. The [Math Review](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-review.pdf) develops counting through successive choices and arrangements. Our practical first step is to describe one complete outcome: a three-person committee, an ordered list of winners, or a code. If you cannot say what makes two outcomes different, a formula can conceal rather than fix the confusion.

For a committee containing Maya, Nikhil and Sara, writing the names in six different orders still describes the same three people. For president, secretary and treasurer, swapping Maya and Nikhil changes who holds each job. The names are identical in both stories; the experiment is not. Words such as “choose” do not automatically mean combinations when the chosen people also receive distinct roles.

Counting questions often arrive as a [Quantitative Comparison](/gre/quant/quantitative-comparison) where the structure, not the arithmetic, decides. Quantity A: the number of three-person groups from ten people. Quantity B: the number of seven-person groups from ten. Choosing three to include is the same as choosing seven to leave out, so the quantities are equal (both 120). Draw labelled slots if order matters and write a set of members if it does not; that small sketch usually settles the model before any arithmetic. The [GRE Quant section guide](/gre/quant) sets out all five question formats.

## Use the smallest counting model that fits

With n distinct objects, an ordered selection of r without repetition has n(n − 1)…(n − r + 1) possibilities. In factorial notation this is n!/(n − r)!. If order is irrelevant, each r-member group appears r! times in that ordered count, so divide again to obtain n!/((n − r)!r!). The division corrects duplicated representations of the same outcome.

Write the short product rather than calculating huge factorials. For eight choose three, use (8 × 7 × 6)/(3 × 2 × 1) = 56\. Computing 8! and 5! separately creates larger numbers without adding reasoning. ETS advises avoiding unnecessary computations and using the calculator when arithmetic is genuinely tedious; cancellation is often more efficient here.

__Counting choices depend on order and repetition__
| Situation                                | Model                                  | Reason                                     |
| ---------------------------------------- | -------------------------------------- | ------------------------------------------ |
| Ordered selection, no repeats            | n!/(n − r)!                            | Later slots have fewer choices             |
| Unordered selection, no repeats          | n!/((n − r)!r!)                        | Divide out arrangements of each group      |
| r slots, n choices each, repeats allowed | n^r                                    | Each slot keeps all n choices              |
| Arrange all distinct objects             | n!                                     | Every position matters                     |
| Arrange repeated types                   | n! divided by repeated-type factorials | Swapping identical objects changes nothing |

## Original problem: compare a committee with roles

Original question: eight students volunteer. How many three-person committees are possible, and how many ways can president, secretary and treasurer be assigned to three different volunteers? The committee count is eight choose three = 56\. The role count is 8 × 7 × 6 = 336, because there are eight choices for president, seven remaining for secretary and six for treasurer.

The ratio 336/56 = 6 is not a coincidence. Every selected three-person committee supports 3! = 6 assignments to the three roles. This is a useful structural check: if your role count equals the committee count, you probably ignored order. If the roles can be held by the same person, however, the no-repeat condition has changed and this particular product is no longer appropriate.

Now impose the original condition that one named volunteer must be on the committee. Choose only the other two members from the remaining seven, giving seven choose two = 21\. The named person is already included, so choosing three from seven would create a four-person committee when that person is added back. Translate “must include” into a fixed member before counting open places.

## Original problem: apply category restrictions

Original question: a group has five chemists and four physicists. A four-person team must contain exactly two chemists and two physicists. Choose the chemists in five choose two = 10 ways, and the physicists in four choose two = 6 ways. Each chemistry pair can combine with each physics pair, giving 10 × 6 = 60 teams.

If the condition changes to at least three chemists, split into disjoint cases. Exactly three chemists and one physicist gives (five choose three)(four choose one) = 10 × 4 = 40\. Four chemists gives five choose four = 5\. The total is 45\. Cases that overlap cannot simply be added; the category counts here make them mutually exclusive.

For at least one physicist, a complement is shorter: all four-person teams number nine choose four = 126\. Teams containing no physicists number five choose four = 5\. Subtracting gives 121\. Compare the case list with the complement before calculating. The same model-choice decision appears in [probability without replacement](/articles/gre-probability-without-replacement), where counts become favourable and total outcomes.

An original small check chooses a two-person team from three engineers and two designers, requiring one of each. The count is 3 × 2 = 6, not five choose two = 10\. The unrestricted count includes three all-engineer pairs and one all-designer pair. Removing those four invalid teams independently reproduces the restricted total of six.

## Original problem: count codes with a leading restriction

Original question: how many four-digit positive integers have different digits? The first position has nine choices, from 1 to 9, because a leading zero would not make a four-digit integer. The second position has nine choices: all ten digits except the first used digit. The last two positions have eight and seven choices. The count is 9 × 9 × 8 × 7 = 4,536.

A four-character access code would be a different experiment if zero were allowed in the first position. With distinct digits it would have 10 × 9 × 8 × 7 = 5,040 possibilities. If repetition were allowed too, it would have 10^4 = 10,000\. A change to the object being counted can matter more than the visible list of digits.

When additional restrictions interact, place the most constrained position first. For a four-digit integer with distinct digits ending in zero, fix zero last, then fill the first three positions in 9 × 8 × 7 = 504 ways. Counting all even-ending cases together needs separate attention to zero versus a nonzero even final digit, because zero changes the available leading choices.

## Original problem: repeated letters and adjacent objects

Original question: how many distinct arrangements use all five letters in LEVEL? There are two Ls, two Es and one V. Treating them as five labelled objects gives 5! arrangements, but swapping the two Ls or the two Es does not change the visible word. Divide by 2! for each repeated type: 5!/(2!2!) = 30.

The correction applies to identical objects, not merely similar ones. Two distinct students with the same first name are still different people if the problem distinguishes them. Conversely, counters identical in colour and otherwise indistinguishable should not acquire artificial labels unless the experiment counts individual objects. Define identity from the question, then apply the repeated-type correction consistently.

A separate original question has five distinct books and asks how many shelf orders keep books A and B adjacent. Treat the pair as one block, leaving four units to arrange in 4! ways. The block itself can be AB or BA, giving 2 × 4! = 48\. If the condition specifies A immediately before B, only AB is allowed and the count is 24.

## Check overcounting with a tiny version

Our recommended check is to shrink the problem. Three volunteers choosing a two-person committee give AB, AC and BC: three groups. The ordered count gives AB, BA, AC, CA, BC and CB: six assignments. Seeing the two representations of each group makes the division by 2! concrete. This check is particularly useful after a restriction makes a familiar formula feel less trustworthy.

Build a practice set that alternates teams, roles, codes and repeated letters. Record whether an error involved order, repetition, a forbidden position or overlapping cases. Once you can choose the model without a label telling you which formula to use, try [topin’s free GRE mock](/gre/practice-test), marked on the official scale, for mixed-topic transfer. This sequence is our study recommendation.

If a question asks which counts satisfy a condition, check every requested choice using the derived model. The [multiple-choice guide](/gre/quant/multiple-choice) covers that interface. Here the important mathematical output is the count of valid outcomes. State it in words before entering it: 60 teams, 336 assignments or 4,536 integers. A naked number can hide a correctly calculated answer to the wrong experiment.

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## FAQs

What is the difference between permutations and combinations?

Permutations count ordered outcomes; combinations count groups whose membership matters but whose internal order does not. Ask whether swapping two selected objects produces a new valid outcome in the experiment.

Does the word choose always mean a combination?

No. Choosing people for distinct roles creates ordered assignments. Define the final outcome from the complete wording instead of relying on a single verb.

Why do combinations divide by r factorial?

An ordered selection represents each r-object group in r! different orders. Dividing removes those duplicated representations, leaving one count for each unordered group.

When should I multiply counts and when should I add them?

Multiply successive compatible choices that together make one outcome. Add disjoint cases that each separately meet the condition. If cases overlap, adjust the overlap or choose another decomposition.

Do I need to calculate large factorials on the GRE?

Usually a shortened product and cancellation are clearer. For n choose r, expand only the r factors needed and divide by r!. Use a calculator for tedious arithmetic after the model is established.

## Sources (checked 5 October 2026)

* [ETS: Quantitative Reasoning overview and calculator guidance (checked 5 October 2026)](https://www.ets.org/gre/test-takers/general-test/prepare/content/quantitative-reasoning.html)
* [ETS: GRE Math Review (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-review.pdf)
* [ETS: GRE Mathematical Conventions (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/gre-math-conventions.pdf)
* [ETS: Guidelines specific to the on-screen calculator (checked 5 October 2026)](https://www.ets.org/content/dam/ets-org/pdfs/gre/on-screen-calculator-guidelines.pdf)

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