# Permutations vs Combinations: One Question Tells You Which

Ask whether swapping two chosen items gives you a different outcome. If yes, it is a permutation. If no, it is a combination. Here is the arithmetic, the factor of r! that separates them, and the two cases neither formula covers.

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- **Canonical URL:** https://dothecalculation.com/blog/math/permutations-vs-combinations
- **Category:** Math
- **Author:** Do The Calculation Team
- **Published:** 2026-08-03
- **Reading time:** 11 min read
- **Publisher:** Do The Calculation (https://dothecalculation.com)
- **Methodology:** https://dothecalculation.com/methodology

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## Permutations vs Combinations

Both count the ways to pick r things from n. They differ on one point: whether rearranging what you picked counts as a new outcome. A permutation says yes, a combination says no, and that single distinction produces every difference between the two formulas.

The test is quick. Pick any two of your chosen items and swap them. If you now have a genuinely different result, order matters and you want a permutation. If nothing has changed, you want a combination.

Tool: [Try the permutation calculator](https://dothecalculation.com/calculators/permutation-calculator) — Enter n and r for ordered selections, with the working shown as a product rather than a raw factorial.

## The two formulas, and how they relate

**Permutations and combinations**

```
P(n, r) = n! / (n − r)!    ·    C(n, r) = n! / ( r! × (n − r)! )
```
- The only difference is the r! in the combination denominator.
- So C(n, r) = P(n, r) / r!, always.
- That r! is the number of ways to rearrange any one selection, which is exactly what the combination is collapsing.

Ten runners, three medals. If you care who takes gold, silver, and bronze, there are P(10, 3) = 10 × 9 × 8 = 720 outcomes. If you only care which three finish in the top three, each set of three has been counted 3! = 6 times, so there are 720 / 6 = 120 outcomes.

**The same three runners, counted six ways**
| Gold | Silver | Bronze |
| --- | --- | --- |
| Ana | Ben | Cleo |
| Ana | Cleo | Ben |
| Ben | Ana | Cleo |
| Ben | Cleo | Ana |
| Cleo | Ana | Ben |
| Cleo | Ben | Ana |

> **What that table shows** — Six distinct permutations, one combination. Every combination corresponds to exactly r! permutations, which is why dividing by r! converts between them and why a combination count is always the smaller of the two.

## There are four cases, not two

Order is only one of the two questions. The other is whether you can pick the same item twice. Together they give four formulas, and most counting mistakes come from ignoring the second question entirely.

**Choosing r from n**
| Order matters? | Repetition allowed? | Formula | Example |
| --- | --- | --- | --- |
| Yes | No | n! / (n − r)! | Gold, silver, bronze from 10 runners |
| Yes | Yes | nʳ | A 4-digit PIN from 10 digits |
| No | No | n! / (r! (n − r)!) | A 3-person committee from 10 people |
| No | Yes | (n + r − 1)! / (r! (n − 1)!) | 3 scoops from 10 ice cream flavours, repeats allowed |

The second row is the one people reach for a permutation formula on and get wrong. A four-digit PIN is 10⁴ = 10,000, not P(10, 4) = 5,040, because 7777 is a legal PIN and the permutation formula forbids it.

The fourth row is the least familiar and the least common. Three scoops from ten flavours where you may repeat gives C(12, 3) = 220, not C(10, 3) = 120, because vanilla-vanilla-chocolate is a valid order.

## Worked examples across all four

**Same n and r, four different answers**
| Question, n = 12, r = 4 | Type | Count |
| --- | --- | --- |
| Four officers with distinct roles | Permutation | 11,880 |
| A four-person committee | Combination | 495 |
| A four-character code from 12 symbols, repeats allowed | Ordered with repetition | 20,736 |
| Four donuts from 12 varieties, repeats allowed | Unordered with repetition | 1,365 |

The officers figure is 12 × 11 × 10 × 9. The committee figure is that divided by 4! = 24. The code figure is 12⁴. The donut figure is C(15, 4). Four plausible readings of the same sentence, and they differ by a factor of forty.

Tool: [Try the combination calculator](https://dothecalculation.com/calculators/combination-calculator) — Enter n and r for unordered selections, useful for committees, hands of cards, and lottery odds.

## Why lottery odds use combinations

A draw of 6 balls from 49 is unordered: matching all six wins regardless of the sequence they came out in. So the count is C(49, 6) = 13,983,816, and a single ticket has one chance in about fourteen million.

The permutation count for the same draw is P(49, 6) = 10,068,347,520, or 720 times larger. If lottery odds were quoted that way, a ticket would look one in ten billion, which is the same event described with a wrong denominator.

> **A useful sanity check** — C(n, r) equals C(n, n − r). Choosing 6 from 49 to include is the same as choosing 43 to exclude. If your two answers do not match, you have made an arithmetic slip somewhere.

## Recognising which one a question wants

**Language cues**
| Wording | Usually means | Why |
| --- | --- | --- |
| Arrange, order, sequence, rank | Permutation | Position is part of the outcome |
| Choose, select, committee, group, subset | Combination | Membership is the outcome |
| President, treasurer, first place | Permutation | Distinct roles make positions different |
| Team, hand of cards, sample | Combination | No role attaches to a position |
| Password, PIN, licence plate | Ordered with repetition | Positions differ and repeats are allowed |
| Handshakes, pairs, edges in a graph | Combination with r = 2 | A shakes B is the same event as B shakes A |

Handshakes is the cleanest small case. Twenty people at a meeting, everyone shakes everyone else's hand once: C(20, 2) = 190. The permutation count of 380 would be right if a handshake had a giver and a receiver, which it does not.

## Passwords are neither, and that is the point

Password strength is an ordered-with-repetition count, which is why it grows so fast. An eight-character password drawn from 62 characters gives 62⁸, about 2.18 × 10¹⁴ possibilities. Adding one character multiplies that by 62; widening the alphabet from 62 characters to 95 multiplies the whole count by about 30, because the widening applies to all eight positions at once.

Neither the permutation nor the combination formula applies here, and using P(62, 8) would give roughly 1.4 × 10¹⁴, understating the space by a third for no reason other than a misapplied formula.

Tool: [Try the password entropy calculator](https://dothecalculation.com/calculators/password-entropy-calculator) — Convert an alphabet size and a length into bits of entropy and a search-space estimate.

## Computing these without overflowing

Do not compute n! and then divide. 49! is a 63-digit number, and most spreadsheets and many languages lose precision or overflow well before that. Cancel the factorials first.

**The cancelled forms**

```
P(n, r) = n × (n−1) × … × (n−r+1)    ·    C(n, r) = that product / r!
```
- P(49, 6) = 49 × 48 × 47 × 46 × 45 × 44, six terms only.
- C(49, 6) divides that by 720.
- In Excel, PERMUT(49,6) and COMBIN(49,6) already do this internally and are safe.
- Use the smaller of r and n − r when computing a combination by hand, since C(49, 43) is the same value with far fewer terms.

## What this does not tell you

- Counting outcomes is not the same as computing a probability. Both formulas assume every outcome is equally likely, which is true for a fair lottery and false for most real situations.
- Neither formula handles items that are indistinguishable from each other. Arranging the letters of a word with repeated letters needs a further division by the factorial of each repeat count.
- Circular arrangements are a separate case. Seating n people around a round table gives (n − 1)! arrangements, not n!, because rotating everyone produces the same seating.
- Constraints break the formulas. "Choose 5 people but at least 2 must be from accounting" cannot be answered by one call to either formula; you split it into cases and add.
- The counts get large quickly enough to be misleading. C(52, 5) = 2,598,960 poker hands sounds tractable and is far beyond enumerating by hand.
- Nothing here says which of the four cases your real problem is. That still requires reading the problem carefully, and it is where nearly all the errors happen.

**What is the quickest way to tell them apart?**

Swap two of the items you picked. If the result is a different outcome, use a permutation. If it is the same outcome, use a combination. Gold and silver swapped is a different podium; two committee members swapped is the same committee.

**Is nPr always bigger than nCr?**

Bigger or equal. They are equal only when r is 0 or 1, since 0! and 1! both equal 1. For every r of 2 or more, the permutation count is r! times larger.

**Why is a 4-digit PIN 10,000 and not 5,040?**

Because digits can repeat. P(10, 4) = 5,040 counts only PINs with four different digits, excluding 1223, 7777, and thousands of others. The right count is 10⁴, one choice from ten for each of four independent positions.

**How do I choose r objects from n when repeats are allowed and order does not matter?**

Use C(n + r − 1, r). For three scoops from ten flavours with repeats, that is C(12, 3) = 220. This case is sometimes called a multiset or a stars-and-bars problem, and it comes up less often than the other three.

**What does the exclamation mark mean?**

Factorial: the product of every whole number from 1 up to that value. 5! = 5 × 4 × 3 × 2 × 1 = 120. By convention 0! = 1, which is what makes the formulas work when r equals n.

**Which Excel functions do these?**

PERMUT(n, r) for ordered without repetition, COMBIN(n, r) for unordered without repetition, PERMUTATIONA(n, r) for ordered with repetition, and COMBINA(n, r) for unordered with repetition. The names map exactly onto the four rows of the table above.

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_Source: [Do The Calculation](https://dothecalculation.com/blog/math/permutations-vs-combinations). Quote freely with attribution and a link to this page._