Count overlapping memberships without counting people twice

Several group lists can legitimately include the same person. Adding their sizes answers a different question from counting distinct identities. Work through this manual set exercise in an existing shared document before labeling a total. The identifiers are invented, and the calculation is not an office attendance report or a claim about new Kiguri users.

Preserve legitimate multiple membership

Define three fictional groups. A contains P1, P2, and P3. B contains P3 and P4. C contains P1, P3, P4, and P5. Treat each identifier as the same invented person wherever it appears.

The lists contain three, two, and four memberships, so their sizes sum to nine. That is a valid membership total. It is not automatically a distinct-person total. Do not remove a legitimate membership to make the lists disjoint, or treat P3 appearing in several groups as an invalid duplicate within one record system.

Build the unique union directly

Collect the identifiers from all three groups and list each only once. The union is P1, P2, P3, P4, P5: five distinct identities.

Keep that explicit list in the review record. It provides an independent check on a calculation based on group sizes. If the supposed same identifier could describe different people, resolve the identity contract before counting. This exercise stipulates stable identifiers, not display-name matching or access to actual member records.

Inspect every overlap

A and B overlap at P3, giving one. A and C overlap at P1 and P3, giving two. B and C overlap at P3 and P4, also giving two. The sum of pairwise overlap counts is five.

Subtracting those five from the membership total nine gives four, which disagrees with the direct union. The error is not in the original group lists. P3 belongs to all three groups, so it was included three times in the total and then subtracted three times in the pairwise correction. Its contribution became zero, even though it is one distinct identity.

Restore the triple intersection once

The intersection of all three groups is P3, with size one. Add that back after subtracting the pairwise overlaps: 9 minus 5 plus 1 equals 5. This agrees with the direct union list.

Write the membership sizes, each pair intersection, the triple intersection, and the final arithmetic beside one another. A reviewer can then point to the identity behind every correction. The add-back is necessary for this three-group calculation; it is not a general instruction to subtract arbitrary overlaps until a desired total appears. A larger collection of groups requires an explicitly appropriate method.

Label the result and test the reasoning

Ask a second participant to explain both nine and five. Nine describes memberships across the three lists; five describes unique stipulated identities. Both can be useful when their meanings are stated. A report calling nine people or reporting four after an incomplete overlap correction would misstate the fixture.

Retain the direct union as well as the arithmetic so a future edit to one list can be checked. The exercise establishes no real participation, customer growth, or availability. Its useful output is an inspectable counting record in which valid overlap is preserved and the reported unit is clear.