Permutation without Repetition Calculator
Count exact ordered selections of r distinct objects chosen from n distinct objects, with no object used twice.
Inputs
Required whole number from 1 to 1,000.
Required whole number from 0 through n.
Results
Generated permutations
All 60 ordered selections for the current startup example.
| No. | Ordered selection |
|---|
How to use this permutation without repetition calculator
What this calculator does
This calculator finds the exact number of ordered selections you can make when you choose r distinct objects from a set of n distinct objects and you cannot choose the same object twice. Order matters: selecting 1, 2, 3 is different from selecting 3, 2, 1. It is useful for counting possible rankings, assignment orders, short codes made from unique symbols, race podiums, and other “choose and arrange” problems. It does not calculate combinations, where order is ignored, and it does not model arrangements that allow repeated choices. For a formal introduction to the rule of product and ordered selections, see the MIT OpenCourseWare notes on counting, permutations, and combinations.
When to use it
Use it when you need to count distinct finishing orders for finalists, assign different people to different ranked roles, arrange a fixed number of unique symbols chosen from a larger set, or verify the size of a sample space before a probability calculation. If the same selected objects in a different order should count as the same outcome, you need a combination instead.
How to calculate
- The calculator opens with the demonstration n = 5 and r = 3. Its result is already calculated, the generator is populated, and Download Excel is immediately available.
- Replace Total number of objects n with the number of distinct available objects. Then set Sample size r to how many positions you will fill. Results update live; no Calculate button is required.
- Read Permutations P(n, r) as the exact count. Use Objects not selected and Multiplication factors to audit the arithmetic. When n is 6 or smaller, inspect the Generated permutations table to see every ordered selection.
- Choose Download Excel to export the current validated inputs, result, supporting details, and generated rows when available. Reset clears the demonstration inputs and results. After Reset, Excel export stays disabled until both required fields again form a complete valid state.
Input guide
Total number of objects n is required and accepts digits only: a whole number from 1 through 1,000, such as 5. It represents how many distinct objects are available before selection begins. Increasing n normally increases the count because each position has more possible choices. Do not enter commas, decimal values, signs, scientific notation, or a count that treats duplicate objects as distinct when they are actually identical.
Sample size r is required and accepts digits only: a whole number from 0 through n, such as 3. It is the number of ordered positions being filled. Raising r multiplies the result by additional descending factors until r reaches n. A value of r = 0 is the standard empty-selection boundary and has one permutation. The most common mistake is entering r greater than n, which is impossible without repetition.
Output guide
Permutations P(n, r) is the exact number of ordered selections. It is an integer identity determined by n and r, not an estimate. Objects not selected equals n – r and helps show why the denominator is (n – r)!. Multiplication factors shows the descending product used to compute the result; long products are compacted for readability while the count remains exact. Permutation generator states whether explicit rows are available. In the Generated permutations table, No. is the row index and Ordered selection is one sequence of distinct labels from 1 through n. The table is intentionally limited to n ≤ 6 so enumeration remains useful rather than overwhelming.
Worked example
The startup example uses n = 5 and r = 3. The first position has 5 choices, the second has 4 remaining choices, and the third has 3. Therefore P(5, 3) = 5 × 4 × 3 = 60. The same result follows from 5! / (5 – 3)! = 120 / 2 = 60. The calculator therefore displays 60 permutations, 2 objects not selected, the factors 5 × 4 × 3, and 60 generated rows. This matches the multiplication-principle treatment of partial permutations in the Discrete Mathematics: An Open Introduction section on combinations and permutations.
Why the formula works
For the first position you may choose any of n objects. Because repetition is prohibited, only n – 1 objects remain for the second position, then n – 2 for the third, and so on. After r positions, the count is the product of exactly r descending integers. Writing the full factorial n! includes extra factors all the way down to 1, so dividing by (n – r)! cancels the factors that belong to unfilled positions. That gives P(n, r) = n!/(n – r)!.
The distinction between permutations and combinations is about whether ordering creates a new outcome. MIT's Counting Rules I lecture places permutations within the broader multiplication-principle framework. In practical terms, a three-person podium is a permutation because gold-silver-bronze order matters, while an unordered three-person committee is a combination.
Boundary cases and common mistakes
When r = n, every object is used and the result is n!, the number of full orderings. When r = 1, the result is simply n. When r = 0, there is one empty ordered selection, which keeps the factorial identity consistent because 0! = 1. Inputs with r greater than n are invalid under a no-repetition model.
Another frequent error is using nr. That expression counts ordered selections when repetition is allowed, because every position keeps all n choices. A no-repetition problem instead loses one available object after each selection. Also avoid dividing by r!: doing so removes ordering distinctions and converts the count toward a combination. If the objects themselves contain indistinguishable duplicates, a different multiset-permutation model is required.