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GRE · Data Analysis · TOPIC LESSON

Counting, permutations, combinations and sets

Choose whether order matters, track overlaps and translate counted outcomes into probabilities only when justified.

GRE General Test45 min study guide10 indexed sections

BY THE END, YOU CAN

  • Apply sum and product counting principles.
  • Distinguish permutations, combinations and repeated-object arrangements.
  • Use Venn regions and inclusion-exclusion without double-counting.

BEFORE YOU START

  • Integer arithmetic and factorial notation
  • Basic probability

Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.

LEARN THROUGH ORIGINAL PRACTICE

Original GRE-style worked examples and practice

All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.

01 / UNDERSTAND & APPLY

Define one outcome precisely before counting

A count is meaningful only after defining when two outcomes are different. If the outcome is a committee, rearranging the same members does not create a new committee. If it is president and treasurer, swapping the two people creates a different assignment. Order means distinct roles or positions, not merely the order in which you happen to make choices.

Separate restrictions such as repetition allowed, items distinct, or at least one category required. A formula selected before these decisions can produce a polished answer to the wrong sample space.

02 / UNDERSTAND & APPLY

Multiply sequential choices; add disjoint alternatives

If a task consists of stages with a fixed number of options at each stage, multiply their counts. If options available later depend on earlier choices, use correct conditional branch counts. Add counts for alternatives only when their outcome sets do not overlap.

For three shirts and four trousers, 3·4=12 outfits if any shirt can pair with any trousers. If a particular pairing is forbidden, subtract it or count allowed branches; do not continue using the unrestricted product.

ORIGINAL WORKED EXAMPLE

Count codes with a first-position restriction

A three-character code uses letters A through E with repetition allowed, but its first character must be A or B. How many codes?

  1. There are 2 choices for the first character.
  2. Each remaining position has 5 choices because repetition is allowed.
  3. Multiply: 2·5·5=50.

Answer 50

03 / UNDERSTAND & APPLY

Distinct ordered selections use falling products

Arranging r distinct objects chosen from n distinct objects without repetition gives n(n−1)…(n−r+1)=n!/(n−r)!. The first position has n choices and each selection removes one. For all n objects, the count is n!. By definition 0!=1, consistent with one empty arrangement.

If repetition is allowed across r positions, the count is nʳ instead. Factorials apply to selections without replacement, not automatically to every ordering problem.

ORIGINAL WORKED EXAMPLE

Assign distinct offices

Seven people are eligible for president, secretary and treasurer, with no person holding two offices. How many assignments?

  1. President has 7 choices. Secretary then has 6, treasurer 5.
  2. The offices are different roles, so order matters.
  3. 7·6·5=210.

Answer 210

04 / UNDERSTAND & APPLY

Remove the internal order when only membership matters

Selecting r members from n without order gives C(n,r)=n!/[r!(n−r)!]. The permutation count overcounts each chosen group r! times, once for each arrangement. Dividing by r! removes that duplication.

C(n,r)=C(n,n−r), since selecting who is included also selects who is excluded. This symmetry can simplify calculations. Do not divide by r! when roles or order actually distinguish outcomes.

ORIGINAL WORKED EXAMPLE

Choose a committee

How many three-person committees can be selected from seven people?

  1. The ordered-selection count is 7·6·5=210.
  2. Each group of three appears 3!=6 times in that count.
  3. 210/6=35 committees.

Answer 35

05 / UNDERSTAND & APPLY

Build restrictions into stages or use a complement

For a committee with a specified composition, choose within each category and multiply: choosing two from one group and one from another gives C(a,2)C(b,1), assuming the groups are disjoint. For at least one from a category, count all groups and subtract those with none.

For arrangements with identical repeated objects, divide n! by the factorial of each repeated multiplicity, because swaps among indistinguishable copies do not produce new arrangements. A problem about actual labeled objects may not permit that division.

ORIGINAL WORKED EXAMPLE

Repeated letters

How many distinct arrangements of the letters in LEVEL are there?

  1. There are 5 positions, with L appearing twice and E twice. V appears once.
  2. Start with 5! arrangements of temporarily labeled copies.
  3. Divide by 2! for the L swaps and 2! for the E swaps: 120/4=30.

Answer 30

06 / UNDERSTAND & APPLY

Fill disjoint regions before adding totals

For two sets, divide the population into A only, both, B only and neither. These four regions are disjoint. Set totals include the overlap, so A only=|A|−|A∩B|. The union counts everyone in at least one set: |A∪B|=|A|+|B|−|A∩B|.

If the total population is N, neither=N−|A∪B|. “Exactly one” is A only plus B only, not the union. A Venn circle’s area is usually schematic; read labeled counts rather than estimating proportions from its drawn size.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Two overlapping activity setsA schematic Venn diagram in a rectangle represents 80 members. Hiking-only region is 27, overlap 18, kayaking-only 14 and neither region 21. Circle areas do not encode counts.HikeKayak271814Neither: 21All members: 80Schematic set regions; areas are not proportional
Region counts, not circle areas, encode the data. Hiking total=27+18=45; kayaking total=18+14=32; union=59; neither=21. The four disjoint regions total 80.

ORIGINAL WORKED EXAMPLE

Count exactly one and neither

Among 80 members, 45 hike, 32 kayak and 18 do both. How many do exactly one, and how many do neither?

  1. Hike only=45−18=27; kayak only=32−18=14.
  2. Exactly one=27+14=41. Union=27+18+14=59.
  3. Neither=80−59=21.

Answer 41 exactly one; 21 neither

07 / UNDERSTAND & APPLY

Three-set inclusion-exclusion corrects the corrections

For three sets, the union equals the sum of the three set sizes minus all three pairwise intersections plus the triple intersection. Pairwise intersection counts include the triple region unless explicitly described as “only.” The triple region was added three times then subtracted three times, so it must be added once.

A region-first diagram is often safer than memorizing a formula. Place the triple count first, subtract it from each inclusive pairwise count to obtain pair-only regions, then fill the single-only regions. Negative resulting counts signal inconsistent data or a mistaken interpretation.

08 / UNDERSTAND & APPLY

Match numerator and denominator to the same outcome type

Counting can calculate probability when all counted outcomes are equally likely. If the denominator counts unordered committees, the numerator must count unordered committees too. Mixing an ordered numerator with an unordered denominator exaggerates the probability.

An original random-selection protocol must justify uniform selection. Choosing a category uniformly and then a person uniformly within it does not select every person equally when category sizes differ.

09 / UNDERSTAND & APPLY

Choose a method and answer the requested question

All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.

A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.

A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.

10 / UNDERSTAND & APPLY

Original practice: order, repetition and overlap

Decide the counting model before computing.

ORIGINAL PRACTICE · 1

Numeric entry: Four desserts and three drinks can be paired freely. How many dessert-drink pairs?

Reveal answer & reasoning

12

Multiply 4·3 because one choice is made at each stage. Adding 4+3 counts individual options, not pairs.

ORIGINAL PRACTICE · 2

Numeric entry: How many four-digit strings use digits 0 through 9 with repetition allowed and leading zero allowed?

Reveal answer & reasoning

10000

Each of four positions has 10 choices, so 10⁴=10000. A string with leading zero allowed is not restricted like a four-digit positive integer.

ORIGINAL PRACTICE · 3

Single answer: How many ways can four distinct books be arranged on a shelf? A 4; B 8; C 12; D 16; E 24.

Reveal answer & reasoning

E

The position choices are 4·3·2·1=24. Repetition is not allowed because each book is used once.

ORIGINAL PRACTICE · 4

Numeric entry: How many two-person groups can be selected from nine people?

Reveal answer & reasoning

36

C(9,2)=9·8/2=36. Dividing by 2 removes the two orders for each pair.

ORIGINAL PRACTICE · 5

Numeric entry: From five designers and four engineers, with no person in both categories, choose two designers and one engineer. How many teams?

Reveal answer & reasoning

40

C(5,2)·C(4,1)=10·4=40. The two designers have no distinct roles, so count their pair without order.

ORIGINAL PRACTICE · 6

Select all that apply: A population has |A|=20, |B|=15, |A∩B|=6 and total 40. Which counts are correct? A A only=14; B union=35; C exactly one=23; D neither=11; E B only=9.

Reveal answer & reasoning

A, C, D and E

A only=14 and B only=9. Union=20+15−6=29; exactly one=14+9=23; neither=40−29=11. B double-counts the intersection.

ORIGINAL PRACTICE · 7

Numeric entry: An unordered two-person group is selected uniformly from five people. What is the probability the group contains a particular person?

Reveal answer & reasoning

2/5

There are C(5,2)=10 equally likely groups. Favorable groups pair the specified person with any of four others, so 4/10=2/5.

ORIGINAL PRACTICE · 8

Numeric entry: Sets A,B,C have sizes 12,13,14; pairwise intersection sizes are 4,5,6; the triple intersection is 2. How large is their union?

Reveal answer & reasoning

26

Inclusion-exclusion gives 12+13+14−4−5−6+2=26. Pairwise counts include the triple region; failing to add it back would undercount.

KEEP THE BIG IDEAS

Quick revision

  • Define an outcome and determine whether roles distinguish order.
  • Multiply stages and add only disjoint alternatives.
  • Combinations remove the order that permutations count.
  • Fill overlap regions carefully and match counting models in probability ratios.
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