Permutations with Repetition – nʳ Arrangements

When each slot can reuse every option, counting multiplies the choices per slot: n to the power r. A 4-digit PIN uses 10 digits for each of 4 positions, giving 10⁴ = 10,000 codes — which is exactly why brute-forcing locks is slow. Choosing 3 times from 2 options yields 2³ = 8 sequences (AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB), all listed by the calculator. The contrast with no-repetition permutations is sharp: P(10,4) without repeats is 10×9×8×7 = 5,040, nearly half the count, because used options vanish from later slots. The tool prints the per-slot choice breakdown, the exponential product, and the full arrangement list for small cases. License plates, toss sequences (2¹⁰ = 1,024 ten-flip runs) and password keyspace estimates all reduce to this single rule.

1. 1

   Enter options n and slots r

   Example: 10 digits and 4 PIN positions

2. 2

   Multiply per-slot choices

   Each of r slots independently offers n options

3. 3

   Compute nʳ

   10⁴ = 10,000 — with the power expansion shown

4. 4

   List small cases

   All 8 sequences appear for n=2, r=3

• 1

  Repetition allowed means exponent, not factorial: nʳ not n!/(n−r)!

• 2

  Each added slot multiplies the count by n again

• 3

  Compare 10⁴ = 10,000 PINs against 5,040 without digit reuse

• Using factorial forms meant for no-repetition setups

• Confusing order: nʳ means r slots each with n choices

• Counting 2³ as 6 by reflex — with repetition it is 8

• Forgetting the empty/all-same sequences are legal outcomes