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PERMUT in Google Sheets: Count Ordered Selections Without Reuse

Statistical Intermediate Google Sheets

Count Sheets permutations with explicit distinct objects, ranked roles, combination comparisons and independent enumeration checks.

Verification status: See the evidence and test scope described in this guide. How formula examples are checked

Quick answer: PERMUT counts ordered selections without reusing a distinct object. With six candidates and three different roles, =PERMUT(6,3) has expected result 120, because the successive choices are 6, 5 and 4.

Availability and version history

  • Introduced: first service launch date is not established by the official history reviewed; no numbered desktop introduction is assigned.
  • Availability: documented in the rolling Google Sheets service. Examples target ordinary web-spreadsheet cells; mobile clients and imported file modes are not separately certified.
  • Checked: October 14, 2026; official PERMUT reference. Documentation and independently calculated examples, without native Sheets execution.

Syntax and the counting model

=PERMUT(n, k)

n is the available pool size and k is the number selected. Google’s reference identifies the standard ordered-selection function and contrasts it with COMBIN, where only the selected subset matters.

Successful examples here use positive integer counts with k<=n. The dedicated page does not establish a complete decimal-truncation, invalid-count or native-error policy. No Excel-specific error or truncation rule is imported into these Sheets examples.

Example 1: enumerate a small control

Suppose the distinct labels are A, B and C, and two different ordered positions must be filled without reuse:

=PERMUT(3,2)

Expected result: 6. The complete list is AB, AC, BA, BC, CA, CB. AB and BA are separate outcomes because they assign the same labels to different positions.

There is no AA outcome: reuse is excluded. If the positions are not distinct and AB is considered the same selection as BA, COMBIN gives 3 instead, representing AB, AC and BC.

Example 2: assign three ranked roles

Enter numeric 6 in A2 and 3 in B2:

=PERMUT(A2,B2)

Expected result: 120. Independently multiply 6×5×4 to check it. Once a candidate occupies one role, that person cannot also occupy another role in this model.

This count assumes every candidate is eligible for every role. If one person cannot take the first role or some combinations are forbidden, the unrestricted PERMUT count is not automatically the answer. Define those restrictions separately.

Example 3: see how the selection length changes the count

For a pool of five distinct objects:

Formula Expected count
=PERMUT(5,1) 5
=PERMUT(5,2) 20
=PERMUT(5,3) 60
=PERMUT(5,4) 120
=PERMUT(5,5) 120

The last two counts agree because, after four positions are filled, only one object remains for the fifth. Equal results for two different formulas can therefore be mathematically correct.

The mathematical empty-selection convention gives one arrangement, but a native k=0 result is not established by the dedicated reference used here and is not part of the certified controls.

Advanced example: split selection from assignment

Choosing any three candidates from six gives 20 unordered groups. Each chosen group has 6 possible orders, because three distinct people can be arranged in 3×2×1 ways.

=COMBIN(6,3)*FACT(3)

Expected result: 120, agreeing with PERMUT(6,3). This decomposition explains why counting groups alone undercounts different role assignments.

A second independent identity is =FACT(6)/FACT(3), again giving 120. These small controls use FACT without claiming that a factorial-ratio implementation is numerically safe for arbitrarily large counts.

Advanced example: reserve a particular person for one role

If a named candidate must take the first of the three roles, choose and order the remaining two people from the other five:

=PERMUT(5,2)

Expected result: 20. If instead that candidate may occupy any one of the three roles, there are three choices for the reserved position, giving =3*PERMUT(5,2) with expected result 60.

These three cases are disjoint because the named person appears exactly once. The unrestricted 120 outcomes divide into 60 containing that person and 60 excluding them; the latter are =PERMUT(5,3).

Reuse, repeated labels and troubleshooting

Allowing each of six choices at every one of three positions gives the different mathematical count 6³=216. PERMUTATIONA addresses reuse, with its own documented limitations and ambiguity notes.

Repeated printed labels can also change what counts as a distinct outcome. PERMUT assumes distinct objects in the pool; it does not automatically collapse indistinguishable labels. Define object identity, order and reuse before selecting a formula.

Use actual integer inputs and inspect the source cells when a count seems implausible. Invalid domains, overflow limits, numeric-text conversion and fractional arguments remain outside the current Sheets evidence rather than being assigned guessed native behavior.

References and verification

Inputs are fictional. Approximate answers are independent mathematical expectations rounded to nine places, not captured native results. Complete type/error matrices, extreme-magnitude accuracy and cross-client behavior remain untested.