Permutations count ordered arrangements; combinations count unordered selections. That single distinction drives every decision in combinatorial mathematics. The fastest way to apply it: the swap test. Take any two chosen items and swap their positions. If the outcome changes, order matters and you have a permutation. If nothing changes, you have a combination. Awarding gold, silver, and bronze medals to three runners is a permutation because swapping two athletes changes who gets which medal. Picking three players for a committee is a combination because swapping two names produces the same group.
Key Takeaways
The single most important rule: permutations count ordered arrangements while combinations count unordered selections, and the swap test tells you which one applies in under five seconds.
| Point | Details |
|---|---|
| Swap test decision rule | Swap two chosen items; if the outcome changes, use a permutation formula. |
| Core formulas | P(n,r) = n!/(n−r)! for permutations; C(n,r) = n!/(r!(n−r)!) for combinations. |
| Repetition changes the formula | Use n^r for permutations with repetition; C(n+r−1, r) for combinations with repetition. |
| Permutations always exceed combinations | For the same n and r (r > 1), permutations produce more outcomes than combinations. |
| Apexapro calculator | Free, browser-based tool at apexapro.com verifies manual calculations instantly, no sign-up needed. |
Table of Contents
- How to tell whether a problem calls for a permutation or a combination
- The core formulas and notation you need to know
- Step-by-step worked examples
- Common mistakes students make and how to avoid them
- Practice problems with answers
- Try the Apexapro permutations and combinations calculator
- An editorial note on learning this material
- The Apexapro calculator is free and ready to use
- Sources
How to tell whether a problem calls for a permutation or a combination
The difference between permutations and combinations comes down to one question: does the arrangement of the selected items matter? Certain words in a problem almost always signal the answer.
Words that signal order matters (permutation):
- arrange, arrangement, order, rank
- first, second, third (or any numbered position)
- schedule, sequence, password, PIN
- seat, place, assign to a role
Words that signal order doesn't matter (combination):
- select, choose, pick, form
- committee, group, team, subset
- sample, collection, hand (as in a card hand)
- lottery, draw
A quick illustration of the swap test in action: suppose you pick letters A, B, and C. As a permutation, ABC and CBA are two different outcomes. As a combination, {A, B, C} and {C, B, A} are the same group. That one mental check resolves most textbook problems in seconds.
Watch for these edge cases:
- Repetition allowed vs. not allowed. A standard lock where you can repeat digits uses a different formula than one where each digit appears only once.
- Multi-stage problems. Sometimes you choose a group first (combination), then arrange it (permutation). Both steps need their own formula.
- Labeled vs. unlabeled positions. Seats in a row are labeled; spots on a generic team are not. Labeled positions almost always mean permutations.
Everyday language is an unreliable guide here. The phrase "combination lock" is a classic misnomer: those locks require numbers entered in a specific order, which makes them permutation locks by strict mathematical definition.
The core formulas and notation you need to know
The formulas follow directly from the swap test logic. According to LibreTexts, the standard formulas are:
- Permutation (no repetition): P(n, r) = n! / (n − r)!
- Combination (no repetition): C(n, r) = n! / (r! × (n − r)!)
The extra r! in the combination formula is the key. When you compute permutations, you count every ordered arrangement. To get combinations, you divide by r! because that is exactly how many ways the r chosen items can be rearranged among themselves. Dividing strips out those redundant orderings, leaving only unique groups.
Repetition cases follow different rules:
- Permutations with repetition: n^r (each of r positions can hold any of n items)
- Combinations with repetition: C(n + r − 1, r)
| Symbol | Meaning | Example |
|---|---|---|
| n | Total number of items to choose from | 10 athletes in a race |
| r | Number of items selected | Choosing 3 positions |
| ! (factorial) | n! = n × (n−1) × ... × 1; 0! = 1 | 4! = 24 |
| nPr or P(n,r) | Permutations of n items taken r at a time | 10P3 = 720 |
| nCr or C(n,r) | Combinations of n items taken r at a time | 10C3 = 120 |
| (n choose r) | Alternate binomial coefficient notation for C(n,r) | (10 choose 3) = 120 |
For any n and r where r is greater than one, permutations produce more outcomes than combinations because permutations consider every ordering distinct while combinations do not.
Step-by-step worked examples
These four problems cover the main scenario types you will encounter on homework and exams.
-
Medals (permutation, no repetition). Ten athletes compete. How many ways can gold, silver, and bronze be awarded?
- Formula: P(10, 3) = 10! / (10 − 3)! = 10! / 7!
- Arithmetic: 10 × 9 × 8 equals seven hundred twenty
- Why permutation? Swapping the gold and silver winners changes the result. Positions are labeled and distinct.
-
Committee (combination, no repetition). From the same 10 athletes, how many 3-person committees can be formed?
- Formula: C(10, 3) = 10! / (3! × 7!)
- Arithmetic: (10 × 9 × 8) divided by (3 × 2 × 1) equals one hundred twenty
- Notice that 720 / 6 = 120: the committee count is exactly the medal count divided by 3!, confirming the derivation. Math is Fun uses this same pair of examples to show why the two formulas differ.
-
Lottery (combination, no repetition). A lottery draws 6 numbers from 1 to 49. Order of the drawn numbers is irrelevant; you win if you match the set.
- Formula: C(49, 6) = 49! / (6! × 43!)
- Arithmetic: the product of 49 through 44 divided by factorial six equals a large number of possible tickets
- Swapping any two drawn numbers leaves the winning set unchanged, so this is a combination.
-
PIN code (permutation with repetition). A 4-digit PIN allows digits 0–9, and repeats are permitted.
- Formula: n^r = 10^4 equals ten thousand possible PINs
- Each position independently holds any of 10 digits, so repetition multiplies the count. A PIN where no digit repeats would use P(10, 4) = 5,040 instead, a notably smaller space.
Common mistakes students make and how to avoid them
The most frequent error is trusting everyday language over the swap test. Words like "arrange" and "choose" seem clear, but problem writers sometimes use them loosely. Always translate the scenario into a concrete question before reaching for a formula.
Typical pitfalls:
- Treating labeled positions as unlabeled. If a problem says "assign a president, vice president, and treasurer," those are three distinct roles. That is a permutation, not a committee selection.
- Forgetting to check for repetition. A password problem where letters can repeat uses n^r, not P(n, r). Missing this produces a drastically wrong answer.
- Double counting in multi-stage problems. When a problem has two stages (choose a team, then pick a captain), apply the combination formula to the team selection and the permutation formula to the captain selection separately, then multiply.
- Misreading "combination" in the problem. As BetterExplained notes, everyday English uses "combination" casually, while mathematics uses it precisely. A "combination of flavors" in a recipe is not necessarily a mathematical combination.
Study tips:
- Apply the swap test before writing any formula. One question, five seconds, correct path.
- Rewrite the problem in positional form: "Is position 1 different from position 2?" If yes, permutation.
- Check whether the problem says "with replacement" or "without replacement." That phrase tells you which repetition formula applies.
Pro Tip: Sketch a small version of the problem with 3 or 4 items and list outcomes by hand. If your list has duplicates (like AB and BA appearing as the same result), you are in combination territory. This scratch-work catches double-counting before it reaches your final answer.
Practice problems with answers
Work through these before checking the answers. Use the swap test on each one first.
-
How many ways can 5 books be arranged on a shelf?
- Answer: P(5, 5) = 5! equals one hundred twenty
- Rationale: Order matters; swapping any two books produces a different arrangement.
-
A coach picks 4 players from a squad of 12 for a relay team where positions are not assigned. How many groups are possible?
- Answer: C(12, 4) = 12! divided by (4! × 8!) equals four hundred ninety-five
- Rationale: No labeled positions; swapping two players leaves the same group.
-
A combination lock has 3 dials, each showing digits 0–9, and digits can repeat. How many codes exist?
- Answer: 10^3 equals one thousand
- Rationale: Permutation with repetition; each dial is independent and order of digits matters.
-
A club of 8 members must choose a 3-person committee, then elect one of those 3 as chair. How many outcomes are possible?
- Answer: C(8, 3) multiplied by P(3, 1) equals one hundred sixty-eight
- Rationale: Two-stage problem. First, form the committee (combination, order irrelevant). Then, select the chair from those 3 (permutation, one labeled role). Multiply the counts.
Try the Apexapro permutations and combinations calculator
The Apexapro permutations and combinations calculator runs directly in your browser, no sign-up and no download required. Enter your values for n and r, toggle between permutation and combination mode, and toggle the repetition switch to see how the count shifts.
To replicate the medal example: set n = 10, r = 3, repetition off, permutation mode. The calculator returns 720, matching the manual arithmetic exactly. Switch to combination mode and the result drops to 120, illustrating in one click why the r! denominator matters.
Try this: After confirming the medal and committee results, change n to 49 and r to 6 in combination mode. The calculator instantly returns 13,983,816, the lottery figure from the worked examples above. Then flip repetition on and watch the count change. That single experiment makes the repetition formulas concrete in a way no textbook page can replicate.
Pro Tip: Use the calculator to check your scratch-work on practice problems, not to skip it. Work the problem by hand first, then verify. Discrepancies tell you exactly where your reasoning went wrong, which is more useful than a correct answer you didn't earn.
An editorial note on learning this material
The swap test was the turning point for making these concepts stick. Before applying it consistently, it was easy to second-guess formula choice on every problem. With it, the decision takes a few seconds and the formula follows automatically.
A focused study routine that works well: solve 10 problems by hand in one sitting, covering at least one permutation, one combination, one repetition case, and one two-stage problem. Then verify every answer with the Apexapro calculator. That cycle of manual work followed by instant verification builds both accuracy and confidence faster than reading examples passively.
The Apexapro calculator is free and ready to use
The Apexapro permutations and combinations calculator gives you instant results for every formula covered in this article, including repetition cases, with no account required.
Head to Apexapro and open the calculator now. Plug in the values from any worked example above, toggle the settings, and see the numbers confirm your manual work. The tool is fully bilingual in English and Spanish, runs on any device, and is updated regularly alongside new math and education tools across the catalog.
Sources
- Difference between Permutations and Combinations - GeeksforGeeks
- 7.2: Permutations and Combinations - Mathematics LibreTexts
- Permutations and combinations - Math is Fun
- Easy Permutations and Combinations – BetterExplained