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COMBINATORICS / PERMUTATION CYCLES

Unsigned Stirling Cycle Calculator

Count permutations of n labelled elements that decompose into exactly k disjoint cycles.

  • 01 Calculated in this tab
  • 02 Values stay in this browser tab
  • 03 Use boundary

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METHOD / WORKED EXAMPLE

Audit the Unsigned Stirling Cycle result step by step

Count permutations of n labelled elements that decompose into exactly k disjoint cycles. The result panel keeps the defining recurrence or counting identity visible so the output can be checked independently.

WORKED DEFAULT

Check the calculation with the default inputs

For n = 7 and k = 3, the recurrence gives c(7,3) = 1,624 permutations with exactly three cycles.

  1. Set rown = 7
  2. Track cyclesk = 3
  3. Apply recurrencec(7,3) = 1,624

READ THE RESULT

Interpret the output in context

The value classifies permutations by disjoint cycle count, not set partitions into unlabeled blocks.

ASSUMPTIONS AND LIMITS

Know where the model stops

  • All n elements are distinct and labelled.
  • The output uses the nonnegative magnitude of the first-kind Stirling number.

n is capped at 18 so every permitted cycle count remains an exact safe integer.

COMMON QUESTIONS

Unsigned Stirling Cycle Calculator FAQs

What definition does Unsigned Stirling Cycle Calculator use?

Build unsigned first-kind Stirling rows from c(n,k) = c(n-1,k-1) + (n-1)c(n-1,k), starting with c(0,0) = 1. The unsigned convention reports a count rather than the alternating sign used by one algebraic notation. Inputs must be whole numbers inside the displayed domain; the page never rounds a decimal into an accepted index. This ties the answer to one explicit convention instead of silently mixing sequence offsets or combinatorial interpretations.

How can I verify the Unsigned Stirling Cycle output?

The value classifies permutations by disjoint cycle count, not set partitions into unlabeled blocks. The previous row supplies the new-cycle and insertion terms. Recompute the displayed recurrence or closed form with the same inputs and compare its previous terms or counting factors. That check supports this bounded result, but it does not transfer the interpretation to a different sequence or counting object.

Why is the Unsigned Stirling Cycle input bounded?

n is capped at 18 so every permitted cycle count remains an exact safe integer. The cap also keeps every intermediate row exact. Combinatorial and recurrence values can grow rapidly even when the inputs look small. The conservative cap prevents browser stalls and avoids presenting an unsafe floating-point integer as exact; larger work needs arbitrary-precision software and independent resource controls.

Use boundary

Calculation path

Build unsigned first-kind Stirling rows from c(n,k) = c(n-1,k-1) + (n-1)c(n-1,k), starting with c(0,0) = 1.

Calculation path

c(n,k) = c(n-1,k-1) + (n-1)c(n-1,k).