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How do you count arrangements and selections using the counting principle, permutations and combinations on the ACT?

Count outcomes with the fundamental counting principle, and distinguish permutations (order matters) from combinations (order does not matter) (Statistics and Probability).

An ACT Statistics answer on counting: the fundamental counting principle, factorials, and telling permutations (order matters) from combinations (order does not matter), with worked ACT-style questions and common traps.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The fundamental counting principle
  3. Permutations: order matters
  4. Combinations: order does not matter
  5. Telling them apart
  6. Why the order question is everything
  7. Try this

What this topic is asking

Counting problems ask how many ways something can happen. The ACT tests the fundamental counting principle (multiply the choices at each stage) and the distinction between permutations (order matters) and combinations (order does not matter). Deciding which applies is the central skill.

The fundamental counting principle

Most ACT counting questions are solved by multiplying.

So with 3 shirts, 2 pants and 4 hats, there are 3×2×4=243 \times 2 \times 4 = 24 outfits. Multiply, do not add, when each stage's choice combines with the others.

Permutations: order matters

A permutation counts ordered arrangements.

Permutations apply to rankings, finishing orders, distinct roles (president then vice-president), and seatings, anywhere swapping two items gives a different outcome.

Combinations: order does not matter

A combination counts selections where order is irrelevant. Choosing 2 people from 6 to form a committee (no roles) is a combination: the pair {Ann, Bob} is the same as {Bob, Ann}. There are 6×52=15\frac{6 \times 5}{2} = 15 such pairs, where the ÷2\div 2 removes the double-counting of the two orders of each pair. In general a combination divides the permutation count by the number of ways to arrange the chosen items, so it is always less than or equal to the matching permutation count.

Telling them apart

The decision rule is simple but essential: ask whether rearranging the same chosen items produces a different result.

  • If yes (order matters), it is a permutation: rankings, sequences, distinct positions.
  • If no (order does not matter), it is a combination: groups, committees, hands of cards, selections.

"Choose a president and a treasurer" is a permutation (the roles differ); "choose two officers" with no specified roles is a combination. Reading for words like "arrange", "order", "rank" (permutation) versus "group", "select", "committee" (combination) usually settles it.

Why the order question is everything

Nearly every counting error comes from choosing the wrong model. The reliable test is the rearrangement question above. For straightforward "how many ways" problems with stages, the counting principle (multiply) is enough. When selecting a subset, decide order-matters (permutation) or order-irrelevant (combination) before computing. Getting that one decision right, then multiplying carefully, secures these points.

Try this

Q1. A password uses 2 letters followed by 1 digit, with repetition allowed (26 letters, 10 digits). How many are possible? [1 point]

  • Cue. 26×26×10=676026 \times 26 \times 10 = 6760 by the counting principle.

Q2. How many ways can 3 finishers (gold, silver, bronze) be chosen from 5 runners? [1 point]

  • Cue. Order matters (distinct medals): 5×4×3=605 \times 4 \times 3 = 60.

Exam-style practice questions

Practice questions written in the style of ACT exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

ACT Math (style)1 marksA diner offers 4 main dishes and 3 desserts. How many different main-and-dessert combinations are possible? (A) 7 (B) 12 (C) 9 (D) 1
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The correct answer is (B), 12.

By the fundamental counting principle, multiply the choices at each stage: 4×3=124 \times 3 = 12 combinations. Choice (A) adds instead of multiplying. Each of the 4 mains can pair with each of the 3 desserts.

ACT Math (style)1 marksIn how many ways can a president and a vice-president be chosen from a club of 6 members? (A) 12 (B) 15 (C) 30 (D) 36
Show worked answer →

The correct answer is (C), 30.

Order matters (president and vice-president are distinct roles), so this is a permutation: 66 choices for president times 55 remaining for vice-president =6×5=30= 6 \times 5 = 30. Choice (B), 15, would be the number of ways to choose 2 members without regard to role (a combination).

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