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Last updated: August 5, 2026

Permutation and Combination Calculator

Quick Answer

Permutations count ordered selections, while combinations count unordered selections. This calculator uses nPr = n!/(n−r)! and nCr = n!/(r!(n−r)!) to compute both values exactly for valid whole-number inputs with 0 ≤ r ≤ n. It is useful for probability, statistics, and everyday counting problems where you need to decide whether order matters.

Use permutations when order matters and combinations when it does not.

Key Takeaways

  • Order matters for permutations and does not matter for combinations.
  • The combination formula removes duplicate orderings by dividing by r!.
  • Exact whole-number arithmetic is best for counting results.
  • Valid inputs require whole numbers with 0 ≤ r ≤ n.
  • Comparing nPr and nCr side by side clarifies the problem type.
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Formula

nPr = n!/(n−r)! and nCr = n!/(r!(n−r)!)

Where:

  • n=total available items
  • r=selected items
  • nPr=ordered arrangements
  • nCr=unordered selections
Permutations and Combinations illustrationA clean teaching diagram for the Permutations and Combinations. It highlights the main variables, places the full formula in a wide banner, and shows a short worked example without overlapping text.Permutations and CombinationsOrderCountCompareInputsRelationshipResultMain relationshipnPr = n!/(n−r)!nCr = n!/(r!(n−r)!)Worked example10P3 = 720, 10C3 = 120Difference = 3! orderings
This illustration summarizes the calculation flow for the Permutations and Combinations: identify the inputs, apply the relationship, and read the supporting outputs that verify the result.

Worked Examples

Committee selection

Choose 3 people from 10 without assigning roles.

  1. 1nCr = 10! / (3! · 7!).
  2. 210 × 9 × 8 ÷ 6 = 120.
  3. 3There are 120 unordered committees.
Final Answer: 120

Podium placements

Assign gold, silver, and bronze from 8 finalists.

  1. 1nPr = 8! / 5! = 8 × 7 × 6.
  2. 2Ordered arrangements total 336.
  3. 3Combinations would be smaller because order would not matter.
Final Answer: 56

Lottery-style selection

Choose 5 numbers from 12.

  1. 1nCr = 12! / (5! · 7!).
  2. 2Compute as (12 × 11 × 10 × 9 × 8) / 120.
  3. 3The number of combinations is 792.
Final Answer: 792

Introduction

Permutation and combination problems ask one central question: does order matter? If the arrangement matters, use permutations. If only the chosen group matters, use combinations. This calculator computes both values from the same inputs, shows the role of r!, and helps you interpret the answer in real contexts such as committees, passwords, awards, schedules, and sampling problems where counting quickly by hand can become cumbersome.

Why order is the deciding idea

Permutations and combinations use the same starting data, but they answer different questions. A committee of three people does not change if you list the names in a different order, so combinations apply. A gold-silver-bronze podium absolutely changes when the order changes, so permutations apply. Recognizing that difference is more important than memorizing formulas, because it tells you which count matches the real situation.

What the formulas mean

The permutation formula nPr = n!/(n−r)! counts ordered selections by multiplying the number of choices at each step. The combination formula nCr = n!/(r!(n−r)!) divides that permutation count by r! because every unordered group can be arranged in r! different orders. In other words, combinations remove duplicate orderings from the permutation total.

How factorials fit in

Factorials compress long descending products such as 10 × 9 × 8 × 7 into a compact notation. They are ideal for counting problems because each choice often reduces the next number of available options by one. The calculator uses exact whole-number arithmetic so large counting results stay reliable rather than being distorted by floating-point rounding.

Worked example

If 10 students are available and you need a 3-person committee, the count is 10C3 = 10!/(3!7!) = 120. If those same students were being assigned president, vice president, and secretary, the count would be 10P3 = 720. The difference between 120 and 720 comes entirely from whether the three chosen people can be ordered in 3! = 6 distinct ways.

Important restrictions on n and r

The values n and r must be whole numbers, and r cannot exceed n. If r is larger than n, the selection request is impossible because you cannot choose more distinct items than exist. Special cases such as r = 0 or r = n are meaningful; there is exactly one way to choose nothing and exactly one way to choose the entire set.

Common errors

The most common error is choosing the wrong formula because the wording of the problem was read too quickly. Another is forgetting that r! appears only in combinations, not in permutations. Students also sometimes treat repeated items as distinct when the problem assumes identical objects, which requires a different counting approach. This calculator is designed for distinct-item selection problems.

Where these counts are used

Permutation and combination counts appear in probability, statistics, coding, genetics, tournament scheduling, quality control sampling, and cryptography. Even outside school, the same reasoning appears whenever you count lineups, rank-order outcomes, or choose subsets from a larger menu of options. Seeing both outputs side by side helps reveal how much order changes the size of a sample space.

Quick sanity checks

For the same n and r, the permutation count should always be at least as large as the combination count because ordering adds possibilities. If r = 1, the two values are identical because selecting one item leaves no ordering issue. If r = n, the combination count becomes 1 while the permutation count becomes n!, which is a helpful reminder that the two questions are fundamentally different.

Quick Reference Card

Permutation and combination cheat sheet

Quick referencePermutation and Combination Calculator

nPr = n!/(n−r)! and nCr = n!/(r!(n−r)!)

Valid range: Whole numbers with 0 ≤ r ≤ n

Common Values

10C3120
10P3720
8P3336
12C5792

Watch Out

  • Do not use combinations when ranking or assigning positions.
  • Do not use permutations when order is irrelevant.
  • r must never exceed n.
  • Factorials grow fast, so expect large exact counts.

Pro Tips

  • Ask “does order matter?” before touching a formula.
  • Use nPr first and divide by r! to get nCr if that feels easier.
  • Special cases such as r = 0 or r = n make good answer checks.
  • If n and r are small, you can often simplify before multiplying everything out.

FAQs

What is the difference between permutation and combination?

Permutation counts ordered arrangements, while combination counts unordered selections.

When should I use nPr?

Use nPr when the order of the chosen items matters, such as races, roles, rankings, or passwords.

When should I use nCr?

Use nCr when only the group matters and different listing orders should count as the same selection.

Can r be bigger than n?

No. You cannot choose more distinct items than are available.

Why does the combination formula divide by r!?

Because each unordered group is counted r! times inside the permutation total, once for every possible ordering.

What happens when r = 0?

Both formulas produce meaningful results: nC0 = 1 and nP0 = 1.

Does this calculator handle repeated items?

It is intended for distinct-item counting. Repeated or identical items require different counting methods.