Count ordered arrangements nPr when rank, sequence, or podium order matters—distinct from combinations where AB equals BA.
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nPr counts ordered arrangements: how many ways to fill r distinct positions from n items when order matters and (usually) items are not reused. 10P3 = 10×9×8 = 720 podium orderings.
10P3 = ?
10 × 9 × 8 = 720 (order matters)
nPr vs nCr?
nPr counts order; nCr ignores it — see comparison below
With repetition?
Use n^r mode when items can repeat (e.g. 10^3 PINs)
Permutations answer ranking problems. Gold, silver, and bronze among 10 sprinters: first place has 10 choices, second has 9, third has 8—10P3 = 720 distinct medal tables where order matters.
Swap two medalists and you get a different permutation. If order were irrelevant, you would use combinations (nCr) instead—same three athletes, one unordered trio.
Formula: nPr = n! / (n−r)! = n × (n−1) × … × (n−r+1)—only r descending factors, not the full factorial. When r = n, nPn = n! (every ordering of all items).
**With repetition** (each slot can reuse items), the count is n^r—different model from standard nPr. A three-digit PIN with digits 0–9 allowing repeats has 10³ = 1,000 codes, not 10P3 = 720.
This calculator uses exact big integers for modest n—critical when homework expects 26P3 = 17,576 exactly, not rounded scientific notation.
Order matters → nPr. Enter n and r for exact nPr, or switch to n^r mode when slots can repeat.
Concise answers for common searches — definitions, steps, and comparisons.
10P3 = 10×9×8 = 720 ordered selections of three items from ten.
Permutation: order matters (ABC ≠ BAC). Combination: order ignored—same letters, one group.
nP0 = 1 for any n ≥ 0. There is exactly one empty arrangement.
When each of r positions can reuse the same item—like a 4-digit PIN with repetition allowed.
Standard mode computes nPr = n!/(n−r)! with BigInt factorials. Repetition mode computes n^r when each of r positions has n choices independently.
Formula
Without repetition: nPr = n!/(n−r)! for 0 ≤ r ≤ n. With repetition: arrangements = n^r. Relation: nCr = nPr / r!.If swapping two chosen items changes the outcome, permutations apply. Committees without roles use combinations.
n is pool size; r is how many ordered slots you fill. Require 0 ≤ r ≤ n in no-replacement mode.
When digits or letters can repeat, use n^r—not standard nPr.
Divide nPr by r! to get combinations when order should not matter.
Input
n = 5, r = 2Output
5P2 = 205×4 = 20 two-letter codes without repeated letters.
Input
n = 10, r = 3Output
10P3 = 720Medal order matters—720 distinct top-three lists.
Input
n = 6, r = 6Output
6P6 = 6! = 720Every ordering of six distinct cards.
Input
n = 9, r = 0Output
9P0 = 1One way to arrange zero items—the empty selection.
Input
n = 26, r = 3Output
26P3 = 17,57626×25×24—no repeated letters in three positions.
Input
n = 10, r = 3, digits repeatOutput
10³ = 1,000 (not 720)PIN model allows 000—permutation without repeat undercounts.
Top three from eight finalists: 8P3 = 8×7×6 = 336 distinct medal orderings.
Three letters from A–Z, no letter twice: 26P3 = 17,576 sequences.
Pick and order 4 people for 4 labeled chairs from 12 candidates: 12P4.
Eight unique characters chosen in order from a charset—permutation with r = 8 when reuse is forbidden.
Step-by-step chains that connect related tools for common tasks.
| Scenario | Tool | 5 pick 2 | Why |
|---|---|---|---|
| President & VP | nPr | 20 | Roles differ |
| Committee of 2 | nCr | 10 | Same pair either order |
| 3-digit PIN, repeat | n^r | 1000 | Slots independent |
| 3-letter code, no repeat | nPr | 60 (from 5) | 5P3 example |
| Notation | Value | Context |
|---|---|---|
| 5P2 | 20 | Intro drills |
| 8P3 | 336 | Sports podium |
| 26P3 | 17,576 | Letter codes |
| nPn | n! | Full permutations |
10P3 = 10×9×8 avoids computing 10! and 7! separately on paper.
nPr ≤ n^r always. Equality only when r = 1 or n is tiny with special cases.
When r = n, verify n! on the factorial calculator.
If problem text says “committee” or “hand,” switch tools before computing.
Medals and permutations of letters in a word need nPr. nCr divides out order.
You cannot pick more distinct items than exist—r must be ≤ n.
Independent slots use n^r. nPr assumes no reuse.
BOOK has duplicate letters—multiset permutations need factorial division by duplicate counts, not raw nPr.
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7P2 = 7×6 = 42. Multiply two descending factors starting at 7.
Standard nPr assumes distinct items. Duplicate letters need adjusted formulas dividing by factorials of repeat counts.
Rotations of a necklace count as one arrangement: (n−1)! for n distinct beads. This tool uses linear nPr.
Because nCr = nPr/r! is always an integer—combinations count unordered subsets.
Yes—empty arrangement convention matches 0! = 1.
If each ordered outcome is equally likely, probability of one specific ordering is 1/nPr.
Exact bigint nPr is supported for n up to 500 in standard mode—sufficient for coursework.
n and r stay in your browser for local calculation.
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Reviewed by EverydayTools Editorial Team on 2026-07-03.
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