Permutation Calculator

Use this Permutation Calculator to enter values, adjust options, and review results in a compact responsive workspace.

Results are calculated automatically as you enter data.

Total items n Chosen items r

Formula and interpretation

Permutation nPr

Disclaimer: All tools on this site are intended for educational and informational purposes only. Estimate only; not professional advice. Verify units, assumptions, precision, and method before relying on results. See our Terms of Use.

Assumptions and limitations

This Permutation Calculator is for general educational use. Results are estimates or mathematical outputs based on the inputs and stated assumptions, not professional engineering, safety, financial, legal, or academic advice. Check units, rounding, formulas, and limitations before using results for important work. For professional, safety-sensitive, construction, engineering, official, or high-cost uses, verify results with an appropriate method, current standards, required tolerances, and a qualified professional. Use of this calculator is subject to our Terms of Use.

▼ See explanations and tips below ▼

Related Calculators

What Are Permutations?

A permutation is an ordered arrangement of items. The key idea is that the same items can form different permutations when they appear in a different order.

For example, the letters A, B, and C can be arranged as ABC, ACB, BAC, BCA, CAB, and CBA. These are different permutations because the order changes.

Permutations are useful when you need to count ordered outcomes without listing every possibility one by one. They are common in probability, schedules, rankings, passwords, race results, seating plans, and role assignments.

The permutation count usually answers a question like: “How many ways can I fill $r$ ordered positions from $n$ available distinct items, without using any item more than once?”

Why Permutations Matter

Permutations help turn a long counting problem into a short calculation. Instead of writing out every possible order, you count how many choices are available at each step and multiply those choices together.

This matters because ordered possibilities can grow very quickly. A small change in $n$ or $r$ can create many more outcomes. For example, choosing three ranked winners from eight competitors gives many more outcomes than simply choosing an unordered group of three competitors.

Permutations are especially helpful when each position has a different meaning, such as first place, second place, and third place; president, vice president, and treasurer; or the first, second, and third characters in a code.

Key Terms to Know

Permutation: An ordered arrangement of selected items.
$n$: The total number of available distinct items or choices.
$r$: The number of ordered positions to fill.
Factorial: The product of all positive whole numbers from a number down to $1$. For example, $5! = 5 \times 4 \times 3 \times 2 \times 1$.
$0!$: Defined as $1$, which makes permutation formulas work when no positions are selected.
Falling product: A product that starts at $n$ and decreases by $1$ for each ordered position.
Combination: An unordered selection. Combinations count groups, while permutations count ordered arrangements.

How Permutations Work

Permutations are based on the multiplication rule for counting. If one task has several possible choices and the next task has fewer remaining choices, the total number of ordered outcomes is found by multiplying the number of choices at each step.

For a permutation without repetition, the first position has $n$ choices. After one item is used, the second position has $n - 1$ choices. The third position has $n - 2$ choices, and the pattern continues until $r$ positions are filled.

The permutation formula is:

$$ {}_{n}P_{r} = \frac{n!}{(n-r)!} $$

Where:

$n$ is the total number of available distinct items.
$r$ is the number of ordered positions being filled.
$n!$ is $n$ factorial.
$(n-r)!$ cancels the unused trailing factors.

The same idea can be written as a falling product:

$$ {}_{n}P_{r} = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1) $$

This falling-product version often makes the meaning easier to see: multiply one factor for each ordered position.

Examples of Permutations in Practice

Example 1: Arranging 3 Items from 8

Suppose there are 8 available items and you want to arrange 3 of them in order. Here, $n = 8$ and $r = 3$.

Using the formula:

$$ {}_{8}P_{3} = \frac{8!}{(8-3)!} = \frac{8!}{5!} $$

Canceling $5!$ leaves the descending product:

$$ {}_{8}P_{3} = 8 \times 7 \times 6 = 336 $$

There are $336$ ordered arrangements.

Example 2: Assigning Different Club Roles

A club has 10 members and needs to choose a president, vice president, secretary, and treasurer. The roles are different, so order matters. Choosing Alex as president and Jordan as treasurer is not the same as choosing Jordan as president and Alex as treasurer.

Here, $n = 10$ and $r = 4$:

$$ {}_{10}P_{4} = 10 \times 9 \times 8 \times 7 = 5040 $$

There are $5{,}040$ ways to assign the four roles.

Example 3: Choosing Zero Ordered Positions

If $r = 0$, there is one way to choose and arrange nothing. This is called the empty product.

For example:

$$ {}_{5}P_{0} = \frac{5!}{(5-0)!} = \frac{5!}{5!} = 1 $$

The result is not zero because there is exactly one empty arrangement.

Permutations vs. Combinations

Permutations and combinations both count selections from a larger set, but they answer different questions.

Situation	Order matters?	Use
Awarding gold, silver, and bronze medals	Yes	Permutation
Choosing three people for an unordered committee	No	Combination
Assigning president, vice president, and treasurer	Yes	Permutation
Selecting three toppings with no order	No	Combination

A combination count for the same $n$ and $r$ is usually smaller than the permutation count because it treats different orders of the same selected items as the same group.

The relationship is:

$$ {}_{n}C_{r} = \frac{{}_{n}P_{r}}{r!} $$

Equivalently:

$$ {}_{n}C_{r} = \frac{n!}{r!(n-r)!} $$

For example, ${}_{8}P_{3} = 336$, but:

$$ {}_{8}C_{3} = \frac{336}{3!} = \frac{336}{6} = 56 $$

The combination result is smaller because each group of 3 distinct items can be ordered in $3! = 6$ different ways.

How to Interpret the Result

The permutation result is the number of ordered arrangements of $r$ distinct positions chosen from $n$ available items.

A larger result means there are more possible orders. This can happen because there are more available items, more positions to fill, or both. Permutation counts often grow quickly because every extra position adds another multiplication factor.

The result is a whole-number count, not a measurement with units. A result of $336$ means there are exactly 336 valid ordered arrangements under the assumptions of the calculation.

When a combination comparison is shown, use it to check whether order is changing the count. For $r = 0$ and $r = 1$, the permutation and combination counts are the same. For most cases where $r \ge 2$, the permutation count is larger because each unordered group can be arranged in more than one order.

If a very large result is shortened in the main display, it has not been rounded. It is only abbreviated for readability. The exact-result display gives the full integer value.

Common Mistakes and Misconceptions

Using permutations when order does not matter. If you are only choosing a group and the order has no meaning, use combinations instead.

Using combinations when order does matter. If each selected item goes into a specific position, rank, role, or sequence, use permutations.

Entering $r$ greater than $n$. For permutations without replacement, you cannot fill more positions than the number of available distinct items.

Trying to use decimals or negative numbers. Permutations in this context count whole items and whole positions, so $n$ and $r$ must be nonnegative whole numbers.

Confusing abbreviation with rounding. A shortened large result is only a display choice. The exact count remains an integer.

Forgetting that $r = 0$ gives $1$. There is one way to arrange no selected items: the empty arrangement.

Using the simple permutation formula for repeated or indistinguishable items. If some items are identical, such as repeated letters in a word, the standard ${}_{n}P_{r}$ formula for distinct items may overcount.

When to Use Permutations

Use permutations when:

The items are distinct.
Order matters.
Items are selected without replacement.
You are filling roles, ranks, slots, positions, or sequence locations.
You need a count of possible ordered outcomes rather than a list of the outcomes.

Common examples include race placements, classroom seating orders, officer elections, ranked prize assignments, ordered codes without repeated characters, and arrangements of selected objects.

Limitations and Things to Keep in Mind

This calculation assumes that all available items are distinct and that no item can be reused after it is chosen. That is why the factors decrease from $n$ to $n-r+1$.

It does not cover permutations with repetition. For example, a code that allows the same digit to appear more than once uses a different counting method.

It also does not cover multiset permutations, where some items are identical. For example, arranging the letters in a word with repeated letters requires adjusting for duplicates.

Inputs must be nonnegative whole numbers. Values such as decimals, negative numbers, comma-formatted numbers, spaces inside the number, plus signs, and scientific notation are not part of this calculator’s supported input format.

The calculator accepts valid inputs up to 500. This is a readability and practicality limit, not a mathematical limit on permutations.

Permutation results are exact whole numbers. No decimal rounding is applied. However, long values may be visually shortened in some result areas, and long falling products may show only the first several factors followed by an ellipsis.

How to Use This Calculator

Enter the total number of available items as $n$.
Enter the number of ordered positions or selected items as $r$.
Make sure both values are whole numbers from $0$ to $500$, and make sure $r \le n$.
Read the ${}_{n}P_{r}$ result as the number of ordered arrangements.
Use the formula substitution, falling product, exact result, and combination comparison to understand how the answer was produced.
Use the example controls, if available, to load sample values such as $n = 8, r = 3$ or $n = 10, r = 4$.
Use the clear control to reset both inputs.

Frequently Asked Questions

What does $nPr$ mean?

$nPr$ means the number of permutations of $n$ distinct items taken $r$ at a time. It counts ordered arrangements, so changing the order creates a different outcome.

What is the difference between $nPr$ and $nCr$?

$nPr$ counts ordered arrangements, while $nCr$ counts unordered selections. If order matters, use $nPr$; if order does not matter, use $nCr$.

Why does the formula divide by $(n-r)!$?

The factorial $n!$ arranges all $n$ items. Dividing by $(n-r)!$ removes the unused part of the full arrangement, leaving only the $r$ positions that matter.

Why is ${}_{n}P_{0} = 1$?

There is exactly one way to choose and arrange nothing. This is the empty arrangement, and it matches the factorial formula because $n! \div n! = 1$.

Can permutations include repeated items?

The basic $nPr$ calculation here assumes distinct items and no replacement. Repeated items or repeated selections require a different method, such as permutations with repetition or multiset permutations.

Why is my result so large?

Permutation counts grow quickly because each ordered position adds another multiplication factor. Even moderate values of $n$ and $r$ can produce very large exact integers.

Is a shortened result rounded?

No. A shortened display is only an abbreviation for readability. The exact result is still a whole-number count and is shown in the exact-result area when available.

Sources and References

Books

Donna Kirk. Contemporary Mathematics. OpenStax, 2023. Chapter 7, Sections 7.1 “The Multiplication Rule for Counting,” 7.2 “Permutations,” and 7.3 “Combinations.” https://openstax.org/books/contemporary-mathematics/pages/7-1-the-multiplication-rule-for-counting, https://openstax.org/books/contemporary-mathematics/pages/7-2-permutations, and https://openstax.org/books/contemporary-mathematics/pages/7-3-combinations. Accessed July 4, 2026.
Jay Abramson. College Algebra 2e. OpenStax, 2021. Chapter 9, Section 9.5 “Counting Principles.” https://openstax.org/books/college-algebra-2e/pages/9-5-counting-principles. Accessed July 4, 2026.