Permutations Calculator (nPr)
Enter n (total items) and r (items to arrange) to calculate the number of permutations P(n, r). A permutation is an ordered selection — the order of chosen items matters. The exact formula is n! / (n−r)!, which this calculator approximates using nʳ. The content below shows how to apply the exact formula with worked examples and step-by-step cancellation.
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Formula
P(n, r) = n! / (n−r)! = n × (n−1) × … × (n−r+1)
P(n, r) counts the number of ways to select r items from n distinct items where order matters. The exact formula is n!/(n−r)!. Writing this out, the (n−r)! terms in the denominator cancel with the matching terms at the bottom of n!, leaving the falling factorial product: n × (n−1) × (n−2) × … × (n−r+1). For P(10, 3): 10 × 9 × 8 = 720. The calculator uses nʳ as an approximation, which equals the exact value only when all n items are being arranged (r = n). For exact results with r < n, use the falling factorial product shown above.
How to use the Permutations Calculator (nPr)
- 1
Enter your n (total items)
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Enter your r (items arranged)
- 3
Read your results instantly
Results update in real time as you type.
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Understanding ordered selection
A permutation answers the question: 'In how many ways can I choose and arrange r items from a pool of n?' The crucial word is 'arrange' — the order of the chosen items matters, so selecting {A, B, C} in the order A-B-C is considered different from B-A-C or C-B-A.
A classic example: a club of 10 members needs to elect a president, vice president, and secretary. The order of assignment matters because each role is distinct. The number of ways to fill these three positions is P(10, 3) = 10 × 9 × 8 = 720. Compare this with the 120 ways to choose a 3-person committee (where roles are identical, so order does not matter).
In general, P(n, r) = C(n, r) × r!, because each unordered group of r can be arranged in r! ways. Permutations always equal or exceed combinations for the same n and r.
The falling factorial product
The most practical way to compute P(n, r) is the falling factorial product: start at n and multiply down through r consecutive integers. P(n, r) = n × (n−1) × (n−2) × … × (n−r+1).
Step through the logic: the first slot can hold any of n items, the second slot any of the remaining n−1, the third any of the remaining n−2, and so on for r slots total. Multiplying these options together gives the total count.
Examples: • P(5, 2) = 5 × 4 = 20 • P(7, 3) = 7 × 6 × 5 = 210 • P(10, 4) = 10 × 9 × 8 × 7 = 5040
This computation never requires evaluating the full factorial n!, making it feasible for larger n.
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Permutations in passwords, codes, and cryptography
Permutation counting underpins security analysis for passwords and combination locks. A 4-digit PIN drawn without repetition from 0–9 has P(10, 4) = 5040 possibilities — far fewer than the 10⁴ = 10,000 possibilities when repetition is allowed, which is the standard PIN model.
In cryptography, permutation ciphers rearrange the letters of a message according to a key. The number of possible keys for an n-letter block is n! — the number of full permutations. For a 26-letter key, 26! ≈ 4 × 10²⁶ is computationally infeasible to brute-force.
Permutation groups are also central to abstract algebra and the mathematics underlying symmetric encryption. Understanding permutation counts is therefore foundational not just for combinatorics homework but for real-world security design.
Tips & Insights
Use the falling factorial, not the full factorials
Computing P(n, r) as n × (n−1) × … × (n−r+1) is faster and less error-prone than computing n! and (n−r)! separately and dividing. For P(100, 3), compute 100 × 99 × 98 = 970,200 directly instead of computing 100! and 97!.
P(n, n) equals n!
When r = n, you are arranging all items, and P(n, n) = n!. This is the 'full permutation' case — the total number of ways to arrange n distinct objects in a sequence.
Distinguish repetition-allowed from without-repetition
Standard permutations assume no item is used twice. If repetition is allowed (e.g., a 3-digit code where digits can repeat), the count is simply nʳ. The calculator's approximation (nʳ) therefore gives the with-repetition answer; use the falling factorial for the without-repetition (standard) case.
Worked Examples
Awarding medals at a race
P(8, 3) = 8 × 7 × 6 = 336. There are 336 ways to assign gold, silver, and bronze medals to runners from a field of 8.
4-digit PIN without repetition
P(10, 4) = 10 × 9 × 8 × 7 = 5040 possible PINs without repeated digits, compared to 10,000 with repetition allowed.
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Frequently Asked Questions
What is the difference between a permutation and a combination?
A permutation is an ordered selection — the arrangement matters. A combination is an unordered selection — only the membership of the group matters. P(n, r) ≥ C(n, r) always, with equality only when r = 0 or r = 1.
How do I compute P(n, r) when n is large?
Use the falling factorial: P(n, r) = n × (n−1) × … × (n−r+1). This requires only r multiplications and avoids computing enormous factorials. For P(100, 3) = 100 × 99 × 98 = 970,200.
What does P(n, 0) equal?
P(n, 0) = 1. There is exactly one way to arrange zero items — the empty arrangement. This matches the formula: n!/(n−0)! = n!/n! = 1.
Are permutations used in probability?
Yes. If all arrangements are equally likely, the probability of a specific ordered outcome is 1/P(n, r). For example, the chance that 3 specific runners finish in a specific order from 8 is 1/P(8, 3) = 1/336 ≈ 0.3%.
What are circular permutations?
In circular arrangements (like seating around a table), one position is fixed to remove rotational equivalence, giving (n−1)! arrangements instead of n!. Two arrangements that differ only by rotation are considered identical.
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