How does the Permutations with Replacement work?

Order matters AND repetition allowed Then: Total = n^k (n choices for each of k positions) Then: Example: 4-digit PIN with digits 0-9: 10^4 = 10,000 possible PINs Then: Contrast: permutations without replacement = n!/(n-k)!.

Can you give an example of how to use the Permutations with Replacement?

Example: Input 3-letter code from 26 letters, repeats allowed gives a result of 26³ = 17,576 codes.

Is the Permutations with Replacement free to use?

Yes — completely free with no registration, download, or subscription required. All calculations happen instantly in your browser.

How accurate is the Permutations with Replacement?

Our Permutations with Replacement uses verified mathematical formulas and is accurate to multiple decimal places. Results are calculated in real-time using the same methods used by professionals.

Matematika

Permutations with Replacement

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P(n,r) — order matters

n (total items)

r (items chosen)

Permutations with replacement count ordered arrangements where items can be repeated. Formula: n^k (n choices, k selections). Used in PIN codes, passwords, combination locks, and license plates.

Type	Formula	n=10, k=4
Permutations with replacement	n^k	10⁴ = 10,000
Permutations without replacement	n!/(n-k)!	10!/6! = 5,040
Combinations without replacement	n!/(k!(n-k)!)	C(10,4) = 210
Combinations with replacement	C(n+k-1,k)	C(13,4) = 715

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Pro Tip

License plate formats illustrate permutations with replacement: a plate with 3 letters + 3 digits has 26³ × 10³ = 17,576,000 combinations — enough for most states' vehicle fleets.

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Did You Know?

A standard combination lock (3 numbers, 0–39) actually uses permutations WITH replacement: 40³ = 64,000 combinations. A determined thief can try all combinations in about 4 hours manually.

References

›Permutation — Wikipedia

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