The Combinatorics of Sibling Conflict
How many ways can n children take sides against each other? The answer is a Stirling number, and it grows exponentially.
Mathematics · A pet project
The Combinatorics of Sibling Conflict
I have four kids, each two years apart, and I grew up with only one sibling. The number of ways they could get into a fight seemed never-ending, so I counted it. The answer for four is 25.
Read the Paper (PDF) Repository
Preprint — queried to The Mathematical Intelligencer on 2026-08-12, reply pending. Not published, not peer-reviewed. Revised since the query, so the PDF here is newer than the one the editors have; the as-queried version is kept in the repository.
The question
A conflict configuration is an unordered pair of disjoint, nonempty subsets of the n children — the two sides — with everyone else uninvolved. That third category is the interesting part: a child can be on side A, on side B, or out of it entirely.
The result
Each child goes to side A, side B, or neither, giving 3n assignments. Drop the ones where a side is empty (2·2n − 1, by inclusion–exclusion), halve it because the labels A and B are interchangeable, and you get:
\[C(n) = \frac{3^n - 2 \cdot 2^n + 1}{2}\]
Which is not an ad hoc expression — it’s S(n+1, 3), a Stirling number of the second kind, OEIS A000392.
The obvious thing to compare it against is Bossard’s 1945 count of family relationships, n(n−1)/2. I had those two side by side as rival quantities until a reader pointed out that they aren’t: a sibling pair is precisely a configuration in which both sides are single children. The pairs don’t run alongside C(n), they sit inside it. Everything else in C(n) is a split where at least one side is a bloc.
| Children n | Conflict configurations C(n) | One against one, n(n−1)/2 |
|---|---|---|
| 1 | 0 | 0 |
| 2 | 1 | 1 |
| 3 | 6 | 3 |
| 4 | 25 | 6 |
| 5 | 90 | 10 |
| 6 | 301 | 15 |
| 7 | 966 | 21 |
| 8 | 3,025 | 28 |
Of the 25 ways four children can take sides, 6 are one against one. Of the 3,025 ways eight children can, 28 are. The containing space grows exponentially; the thin slice of it where nobody sides with anybody grows quadratically.
What it doesn’t claim
It counts a possibility space — not how often siblings actually fight, how badly, or how any of it turns out. Plenty of big families are close and plenty of small ones aren’t.
It’s also not a novelty claim over the field. Bossard (1945) counted family relationships and Kephart (1950) extended that to subgroupings; the paper claims only to extend that line to opposing-sides configurations, and to say plainly that Bossard’s count sits inside the new one rather than beside it.
The counts were verified two ways — brute-force enumeration of all 3n assignments for n = 1–8, and the closed form against OEIS. The figures went through four rounds of blind reading panels; the third is what caught the subset error above. The records are in the repository.
Links
- The Paper (PDF) — four figures and one table, including two worked examples not shown here
- Repository — manuscript source, figure build scripts, chart review and panel returns
- OEIS A000392 — the sequence, S(n+1, 3)