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A permutation is an arrangement of items where the order matters. The password 4729 and the password 9742 use the same digits but are different permutations — type one into the other’s login box and nothing happens. That order-sensitivity is the entire difference between permutations and combinations: permutations count arrangements, combinations count selections.
The counting logic is the fundamental counting principle: if the first choice can be made in n ways, the second in n−1 ways (one item is used up), the third in n−2 ways, and so on, then the total number of ordered arrangements of r items from n is the product n × (n−1) × ··· × (n−r+1). No repetition is allowed — once an item is placed, it cannot be picked again.
When r equals n, you are arranging everything, and the product runs all the way down to 1: that is the factorial, n!. The permutation formula nPr = n!/(n−r)! is really just a shorthand that cancels the unused tail of the factorial. For n = 10, r = 3: 10!/7! = (10×9×8×7!)/7! = 10×9×8 = 720, and the 7! cancels exactly as the formula promises.
Permutations show up anywhere order is meaningful: race finishes (who got gold, silver, bronze), passwords and PINs, batting orders, seating charts, and the number of ways to shuffle a deck. Whenever swapping two items changes the outcome, you are counting permutations.
Example 1 — podium finishes. Twelve runners compete; gold, silver and bronze medals are awarded. How many podium outcomes are possible? Order matters (gold vs. bronze is a different result), so this is P(12, 3) = 12 × 11 × 10 = 1,320. Note that choosing which three runners medal (without order) would be far fewer — that is a combination, not a permutation.
Example 2 — a 4-digit PIN. Digits 0–9, no repeats allowed, order matters. That is P(10, 4) = 10 × 9 × 8 × 7 = 5,040 possible PINs. This is why banks forbid repeated digits in some systems — allowing repeats would jump the count to 10&sup4; = 10,000, but it would also let people pick 1111.
Example 3 — arranging books. You own 7 distinct novels and want to arrange all of them on a shelf. That is P(7, 7) = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 orders. The calculator’s step display walks through this product so you can verify each multiplication by hand.
A website requires an 8-character password using 26 lowercase letters, with no repeated letters. How large is the search space an attacker faces? That is P(26, 8) = 26 × 25 × ··· × 19 = 62,990,928,000 — about 63 billion possibilities. At a million guesses per second, brute force would take roughly 17.5 hours. Add uppercase letters, digits and symbols (say 70 distinct characters) and P(70, 8) explodes past 10¹&sup4;, which is why length plus a large alphabet is the real defense. Every password-strength meter on the internet is, at heart, a permutation calculator with this formula.
A quiz league has 9 teams and each episode features 3 teams competing in fixed roles — first challenger, second challenger, and defending champion. How many distinct episode lineups are possible? Roles make order matter, so it is P(9, 3) = 9 × 8 × 7 = 504 lineups. If instead the three teams just played a round-robin with no distinguished roles, the answer would be C(9, 3) = 84 — exactly 6 times smaller, because each unordered trio can be assigned to the 3 roles in 3! = 6 ways. That factor of r! is precisely the relationship nPr = r! × nCr, and it is the fastest way to sanity-check a combinatorics answer.
Three sibling formulas cover nearly all basic counting, and choosing the right one is the whole skill. Permutations (nPr): order matters, no repetition — race results, passwords without repeats. Combinations (nCr): order does not matter, no repetition — lottery draws, committees; nCr = nPr / r!. Variations with repetition (nr): order matters, repetition allowed — PINs that permit 1111, license plates. A quick decision rule: ask ‘if I swap two chosen items, does the outcome change?’ If yes, it is a permutation (or nr if repeats are allowed); if no, it is a combination. Our Factorial Calculator handles the n! building block directly when you need it.
The nPr formula assumes every item is used at most once, but real problems sometimes allow repeats. A 4-digit PIN that permits 1111 is not P(10, 4) — it is 104 = 10,000, because each of the 4 positions independently offers 10 choices. The general rule for permutations with repetition is nr: n choices for each of r positions. License plates with 3 letters followed by 3 digits give 263 × 103 = 17,576,000 combinations — and notice how allowing repetition explodes the count compared with the no-repeat version.
A second twist is the circular permutation: seating n people around a round table. Rotating everyone one seat clockwise changes nobody’s neighbors, so the n rotations of any seating are the same arrangement. The count is (n−1)! instead of n! — fix one person’s seat to break the symmetry, then arrange the rest linearly. For 8 dinner guests that is 7! = 5,040 seatings rather than 40,320. If flipping the table over is also considered the same (as with a necklace), divide by 2 more.
A third variant: permutations of a multiset, where some items are identical. The word BANANA has 6 letters but only 60 distinct arrangements, not 720, because the three A’s and two N’s are interchangeable. The formula divides by the factorial of each repeat count: 6!/(3!×2!) = 60. Anagram puzzles and Scrabble racks live here — our Even Permutations Calculator and Odd Permutations Calculator go deeper on the structure of individual arrangements, including parity and inversion counts.
When you meet a new counting problem, run this checklist: (1) does order matter? (2) is repetition allowed? (3) are all items distinct, or are some identical? (4) is the arrangement linear or circular? The answers pick the formula — nPr, nCr, nr, (n−1)!, or the multiset division — and nine out of ten combinatorics errors come from skipping this checklist rather than from arithmetic.
It helps to have a feel for the scale. P(5, 2) = 20 — small enough to list by hand. P(10, 3) = 720, the classic textbook example. P(26, 5) = 7,893,600 possible 5-letter strings from the alphabet. P(52, 5) = 311,875,200 — the number of ordered 5-card deals from a deck, which is why poker hand rankings need the combination count instead (2,598,960 unordered hands). And P(100, 3) = 970,200, a reminder that even choosing just 3 from 100 explodes into the hundreds of thousands. When a result looks absurdly large, that is usually the math telling you the truth: ordered counting grows brutally fast, which is precisely why passwords work.
Factorial Calculator computes n! directly — the building block inside every permutation formula.
Odd Permutations Calculator analyzes the parity of a permutation — whether it decomposes into an odd or even number of swaps.
Even Permutations Calculator the companion tool for even permutations, with inversion counting.
Pascal’s Triangle Calculator generates binomial coefficients C(n, k) — the combination numbers that relate to permutations by nCr = nPr / r!.
A permutation counts ordered arrangements without repetition: nPr = n!/(n−r)! = n×(n−1)×···×(n−r+1). Use permutations when swapping two chosen items changes the outcome (passwords, rankings); use combinations when it does not. Relationship: nPr = r! × nCr. Special cases: nP0 = 1, nPn = n!.
nPr = n! / (n − r)!, where n is the total number of items and r is how many you arrange. It equals the product n × (n−1) × ··· × (n−r+1).
720. It is 10 × 9 × 8 = 720 — the number of ordered triples you can form from 10 distinct items.
Use permutations when order matters: swapping two chosen items must change the outcome (passwords, race finishes, seating order). If order is irrelevant (committees, lottery numbers), use combinations instead.
nPr counts ordered arrangements; nCr counts unordered selections. They are related by nPr = r! × nCr, because each unordered selection of r items can be ordered in r! ways.
No. You cannot arrange more items than exist, so r ≤ n is required. If r = n, the permutation is just n! (arrange everything).
0! = 1 by definition. It makes the edge cases consistent: nP0 = n!/n! = 1 (one way to arrange nothing) and the combination formula works when r = 0 or r = n.
P(26, 5) = 26 × 25 × 24 × 23 × 22 = 7,893,600. Real words are a tiny subset — most of these are unpronounceable strings.
Factorials grow explosively: 20! already exceeds 2 × 10^18. Always cancel before multiplying (compute 100!/97! as 100 × 99 × 98) instead of evaluating each factorial separately.
Cite this page as: Sigma Calculator. “Permutation Calculator.” https://sigmacalculator.net/math-calculators/permutation-calculator/. Accessed October 12, 2026.
Last updated: October 12, 2026