Four questions to ask before calculating
ETS places counting in the Data Analysis part of Quantitative Reasoning, alongside descriptive statistics and probability; it includes combinations, permutations, and Venn diagrams. Quantitative Reasoning has 27 questions across two sections, with 47 minutes in total and a basic on-screen calculator. This article uses two formats: Quantitative Comparison and Numeric Entry, where you enter an integer or decimal in one box, or a fraction in two boxes.
Use the same questions, in this order, to solve any counting problem. ETS’s Math Review starts with the key distinction: in a list, order and repetition matter; in a set, neither does.
- What are you counting: sequences of choices, groups, or the elements of a set?
- Does order change the result? If swapping two elements creates a different outcome, order matters.
- Can the same item be chosen more than once?
- Do any sets overlap, or does a condition rule out some cases?
The multiplication principle, including repetition
If a first choice has k possibilities and an independent second choice has m, there are k × m possibilities altogether; with more independent choices, multiply all the counts. When repetition is not allowed, the choices are no longer independent, but the method stays the same: at each step, count what has not yet been chosen.
Towns X, Y and Z lie along one line. Six two-way roads join X to Y, and three two-way roads join Y to Z. A courier drives from X to Z and then back to X, passing through Y each way, and never uses the same road twice. How many different round trips are possible?
Enter your answer before viewing the review.
The solution, step by step
180.
The method in action
- Outbound: 6 roads from X to Y, then 3 from Y to Z. On the return trip, exclude the Y–Z road already used (2 of 3 choices), then the X–Y road already used (5 of 6 choices). Total: 6 × 3 × 2 × 5 = 180.
- Check: two distinct X–Y roads in order give 6 × 5 = 30; two Y–Z roads give 3 × 2 = 6; 30 × 6 = 180. The answer 324 (6 × 3 × 3 × 6) would allow roads to be reused.
Arrange or simply choose
Arranging n distinct objects gives n! = n × (n − 1) × … × 1 possibilities. Choosing k of those n objects without repetition, where 0 ≤ k ≤ n, and arranging them gives n!/(n − k)! permutations. Choosing the same k objects without regard to order gives n!/(k! × (n − k)!) combinations, because each group has been counted k! times, once for each possible order. For example, assigning three distinct roles among 8 volunteers gives 8 × 7 × 6 = 336 assignments; choosing a committee of 3 with no roles gives only 336/3! = 56 groups. ETS also notes the convention 0! = 1 and that “n choose k” equals “n choose n − k.”
Simplify factorials before calculating: 10!/(3! × 7!) becomes (10 × 9 × 8)/(3 × 2 × 1). ETS advises against using the calculator for a simple calculation that is faster to do mentally.
Two sets: subtract the intersection
For two finite sets, the number of elements in their union is the sum of the two set sizes minus the size of their intersection: |A ∪ B| = |A| + |B| − |A ∩ B|, where the bars denote the number of elements. Subtracting the intersection prevents double-counting, as ETS explains in its Math Review. Two other relationships follow: the total equals the size of the union plus the number of elements outside both circles, and “exactly one of the two” equals |A| + |B| − 2 × |A ∩ B|.
The four non-overlapping regions are A only, B only, both, and neither. “Exactly one” combines the first two; “at least one” combines the first three.
Among 60 students, 35 study German and 28 study Spanish. Nine of the 60 study neither language. How many of the students study exactly one of the two languages?
Enter your answer before viewing the review.
The solution, step by step
39.
The method in action
- At least one language: 60 − 9 = 51. So 35 + 28 − (both) = 51, which means 12 students study both languages. Exactly one language: 51 − 12 = 39.
- Another approach: German only is 35 − 12 = 23, and Spanish only is 28 − 12 = 16; 23 + 16 = 39. Common errors are 51 (forgetting “exactly”) and 63 (adding without subtracting the overlap, which exceeds the total number of students).
In a class of 30 students, 18 play chess and 15 play go. Quantity A: the number of students who play both games. Quantity B: 3. Compare the two quantities.
Choose an answer.
The solution, step by step
D.
The method in action
- The prompt does not say whether every student plays at least one game. If so, 18 + 15 − (both) = 30 gives 3 students who play both: A = B. If 7 students play neither, then 18 + 15 − 10 = 23 students play at least one game, and 23 + 7 = 30: A = 10, which is greater than B.
- Two valid situations produce two different relationships, so the answer is D. The calculation 18 + 15 − 30 = 3 gives only a minimum. Choosing C assumes every student plays at least one game; choosing A assumes some students play neither. The prompt establishes neither assumption.
Three sets: count region by region
ETS’s Math Review states the principle for two sets and illustrates three with a diagram. For three sets, the safest method is to fill the seven regions and the outside area from the center outward: first the members in all three, then each pair minus that center, then each set alone. A pair count includes the center when the prompt says “both” without further detail; read carefully to check.
The equivalent formula is the sum of the three set sizes, minus the three pair counts, plus the center. Use the regions as a check: none should be negative, and their sum should not exceed the total population.
A club has 75 members. Of them, 30 run, 26 swim and 22 do yoga. Also, 10 members both run and swim, 8 both run and do yoga, and 7 both swim and do yoga; each of these three counts includes the members who do all three activities. Exactly 4 members do all three activities. How many members do none of the three activities?
Enter your answer before viewing the review.
The solution, step by step
18.
The method in action
- From the center outward: all three activities, 4; running and swimming only, 10 − 4 = 6; running and yoga only, 8 − 4 = 4; swimming and yoga only, 7 − 4 = 3; running only, 30 − 6 − 4 − 4 = 16; swimming only, 26 − 6 − 3 − 4 = 13; yoga only, 22 − 4 − 3 − 4 = 11. The seven regions total 57, so 75 − 57 = 18.
- Check: 30 + 26 + 22 − (10 + 8 + 7) + 4 = 57. Forgetting to add the 4 back gives 53 and an answer of 22; failing to subtract any pair gives 78, more than the total of 75.
Count what is forbidden
When a condition is expressed as a negation (“not adjacent,” “at least one,” “none”), it is often easier to count the opposite: total cases minus forbidden cases. Check that the two groups do not overlap and that their counts add up to the total.
Five people, including Ana and Ben, take seats in a row of six chairs, so that exactly one chair stays empty. In how many different arrangements are Ana and Ben not in adjacent chairs?
Enter your answer before viewing the review.
The solution, step by step
480.
The method in action
- All arrangements: the five people choose seats in turn, giving 6 × 5 × 4 × 3 × 2 = 720. Ana and Ben in adjacent seats: 5 adjacent pairs of chairs and 2 orders, or 10 placements; the other three then choose 3 of the 4 remaining chairs, 4 × 3 × 2 = 24. Forbidden cases: 10 × 24 = 240. Answer: 720 − 240 = 480.
- Direct check: if Ana takes an end seat (2 cases), Ben has 4 non-adjacent seats; if she takes an interior seat (4 cases), he has 3. That gives 2 × 4 + 4 × 3 = 20 placements, then 20 × 24 = 480. Forgetting the 2 orders for Ana and Ben gives 720 − 120 = 600; forgetting the empty chair and seating five people in five chairs gives 120 − 48 = 72.
The pitfall
Double-counting, forgetting order, or accepting an impossible case
Three errors are common. Double-counting: adding two set sizes without subtracting their intersection, or interpreting “10 do both” as “10 do only both.” Forgetting order: treating distinct roles as a group (56 instead of 336), or counting arrangements when only the group matters. And impossible cases: a negative region or one larger than its set almost always means you misread a piece of information.
In a Quantitative Comparison, do not choose D simply because the counting is difficult: you need two valid situations that produce different relationships.
From counting to probability
When all outcomes are equally likely, probability equals the number of favorable cases divided by the number of possible cases; count both using the same rule for order. For example, choose 2 people at random from 5, of whom 2 are women. The probability of choosing both women is 1/10, because one of the ten possible pairs is favorable.
ETS also includes compound, independent, and conditional probability under Data Analysis; see the article on statistics and probability for more.
Key takeaways
- Before calculating, ask four questions: what are you counting, does order matter, is repetition allowed, and do any sets overlap?
- For k distinct objects chosen without repetition from n, a combination is a permutation divided by the k! equivalent arrangements: n!/(k! × (n − k)!).
- Two sets: A + B − (A and B). Three sets: fill the regions from the center outward and check their sum.
- A case described by a negation can often be counted using its complement: total minus forbidden cases.