Mara builds a noncontainment collection from four labels

Mara begins an invented tabletop project with four labels and a question about a noncontainment collection. She wants several choice cards, each naming some of those labels, with no selected card containing all the labels of another selected card. Her finished maximum is six cards. They are the six different two-label choices. A small counting argument explains why adding a seventh cannot work under her rule.

Mara and the cards belong to this fictional exercise. The labels do not describe actual products, people’s preferences or compatible equipment. She calls them A, B, C and D so their membership can be checked without extra associations. A card is simply a subset of those four labels. It contains each chosen label once, and the order in which the labels are written does not create another card.

Put the collection rule beside the blank cards

For example, AB means the subset containing A and B. BA is the same subset, so Mara keeps only one record for it. ABC contains every member of AB and one additional label, C. Those two cards cannot both enter her collection. The rule concerns containment between distinct selected subsets. It does not say that a card must contain exactly two labels; that size emerges later from the maximum argument.

Overlap is permitted. AB and AC share A, but AB lacks C and AC lacks B. Neither contains the other, so that pair passes the rule. Requiring no shared labels would be a different, stricter exercise. Mara writes this comparison first because otherwise a useful collection could be rejected merely for reusing a label, even though reuse is allowed by the actual condition.

There are sixteen possible cards before the collection rule is applied. Each of the four labels can be present or absent, giving two choices at each position. The empty subset and the full ABCD subset are included in this inventory. A collection can select any number of different cards, but must pass the containment test for every pair it selects. Repeating the same card cannot increase its size.

Lay out the six-card witness

Mara writes AB, AC, AD, BC, BD and CD. These are all the two-label subsets. To count them without a duplicate, she can choose a first label in four ways and a different second label in three ways, then divide by two because AB and BA describe the same pair. The result is six. Every pair of distinct displayed cards has two members, so neither can properly contain the other.

The last statement can be checked directly. If one two-member set contained another two-member set, both members of the second would already fill the first. They would then be the same set, contrary to selecting distinct cards. Thus the six-card display is an allowed collection. It supplies a lower bound: the largest possible collection has at least six cards. It does not yet prove that six is the maximum.

Mara leaves a seventh blank card nearby to keep the unanswered part visible. Failing to find a label combination for that card would not prove that no alternative seven-card collection exists. Perhaps another arrangement of differently sized subsets could do better. The next task is an upper bound that applies to every allowed collection, rather than another inspection of her preferred six-card display.

Try a mixed-size collection without changing the rule

The collection A, BC, BD and CD is allowed. None of its three pairs contains A, and the singleton A cannot contain either member of a pair. The three distinct pair cards do not contain one another. This four-card example shows that mixed sizes are permitted. A rule requiring equal card sizes would incorrectly discard it, even though it is smaller than the proposed maximum.

Now try to add ABC to the six-pair display. It contains AB, AC and BC, so it conflicts with several cards already selected. Adding singleton A also fails, since AB, AC and AD each contain it. These checks explain why obvious additions to the six-card witness fail. They remain examples, not the universal upper-bound proof: another seven-card proposal might omit some of those conflicting pairs.

The two boundary cards are especially restrictive. The empty set is contained in every subset, so a collection containing it can contain no other distinct card. The full set ABCD contains every other subset, so it too can appear only in a one-card collection. The boundary cards are allowed inputs, but their consequences must be retained. They are not omitted from the sixteen-card inventory to make the final count easier.

Make a nested chain from an ordering of the labels

Mara next orders the four different labels. An ordering such as B, A, D, C generates five nested subsets: empty, B, AB, ABD and ABCD. Start with nothing and add one label at each step, following that ordering. Each subset contains the preceding one. The middle AB is written in a standard label order for readability; it records membership, not the order in which A and B arrived.

There are twenty-four different label orderings: four choices for the first label, three for the next, two for the next and one for the last. Each generates a chain of this form. Different orderings generate different full chains because the successive additions recover the ordering. The chains may share individual subsets, including their empty and full endpoints. Sharing a subset does not make two whole chains identical.

Any allowed noncontainment collection meets each one of these nested chains in at most one card. If it selected two different cards from a chain, the later one in that chain would contain the earlier one. That would violate the original rule. This is the structural link between the card project and the twenty-four chains. It applies to every allowed collection, regardless of its card sizes.

Count how often one card appears among the chains

Take a particular pair card, AB. A chain contains AB exactly when its first two added labels are A and B in either order. That prefix has two possible orders. The remaining C and D can then be added in either order, giving two choices for the suffix. Multiplying gives four chains containing AB. The same argument holds for any of the six pair cards.

A singleton such as A lies in six chains. A must be the first addition, and the remaining three labels can be ordered in six ways. A triple such as ABC also lies in six chains: its three labels occupy the first three positions in any of their six orders, and D is last. Thus a singleton or triple uses more of the available chains than a pair card does.

The empty card appears at the start of all twenty-four chains. ABCD appears at the end of all twenty-four. We have now accounted for every possible card size: zero and four labels give twenty-four chains each; one and three give six each; two give four. Every card appears in at least four chains. Equality occurs precisely for the two-label cards.

One compact way to retain the calculation is to let k be a card’s number of labels. Its k members must come first in the ordering, in any of k factorial orders. The remaining four minus k labels come afterward, in any of their factorial orders. The product counts the chains containing that card. At the boundary, zero factorial is one, consistent with the single empty ordering for an empty block.

Count the card-and-chain records in two ways

Suppose an allowed collection has m cards. Record a pair whenever a selected card belongs to one of the twenty-four chains. Counting by selected cards gives at least four records for each card, hence at least four times m records. A card appearing in several chains legitimately contributes several different records. The objects being counted are card-and-chain pairs, rather than just the selected cards.

Count those same records by chains instead. Every chain contains at most one selected card, so each of the twenty-four chains contributes at most one record. The total number of records is at most twenty-four. Combining the two counts gives four times m at most twenty-four, and dividing by four gives m at most six. This upper bound covers any choice of card sizes and memberships.

The six-pair display already established that six is achievable. The universal upper bound now says that no allowed collection can exceed six. Together they prove the exact maximum. Mara can remove the seventh blank from the proposed display while retaining the reason it cannot be filled as part of any seven-card solution. The conclusion follows from the counting records, not from the appearance of her layout.

See what equality says about the finished collection

If an allowed collection has six cards, its cards generate at least twenty-four card-and-chain records. Since there can be at most twenty-four, equality must hold. Every selected card must therefore belong to exactly four chains. A singleton, triple, empty card or full card would contribute more and push the count above twenty-four, because each of the other five cards still contributes at least four.

Only pair cards occur in exactly four chains. Consequently every card in a six-card solution has two labels. There are exactly six such cards available, and the collection has no duplicates, so it must contain all six. This proves uniqueness of the maximizing set collection on the fixed labels A, B, C and D. It does not force a particular order for placing the six cards on a table.

That distinction leaves Mara room to arrange her display. She can group cards beginning with A or place overlapping pairs beside one another, provided every member remains visible. The mathematical object is the collection of subsets, not a preferred physical layout. Moving AB from the left to the right creates no new maximizing collection. Renaming the four abstract labels consistently changes their notation while preserving the stated membership relationships.

Audit the finite controls and preserve the limits

An exact control enumerates all sixteen subsets and all sixty-five thousand five hundred thirty-six possible selections of those cards. Exactly one hundred sixty-eight selections satisfy the noncontainment rule, including the empty collection. The largest have six cards, and there is exactly one maximizing selection: the six pairs. A separate control builds all twenty-four permutation chains and checks every subset’s chain count.

These checks cover the complete defined four-label exercise. The chain argument supplies a short proof that can be read without rerunning that enumeration. The article does not establish a general maximum formula for every number of labels or for collections with additional restrictions. If a later project forbids a pair, allows duplicate cards or replaces containment with disjointness, it needs its own inventory and argument.

The counter allocation worksheet minimizes a squared-count score by transfers. Mara’s project instead maximizes a collection size under an inclusion rule. A balanced count pattern cannot replace the chain-incidence upper bound.

The mixed strategy security example joins an achievable result with an opposing bound to prove its value. This project also retains a witness and bound, while using deterministic subsets and chains rather than probabilities or payoffs.

Keep the optional native accounts separate from the cards

The Zeckos tuna weather-vane listing has three images of a fish above an arrow, cardinal cross, pole, balls and dedicated roof bracket. Images do not authenticate material, finish, dimensions, weather resistance or directional accuracy. Its mounting components serve that host arrangement, without supplying the mathematical card inventory.

The Hidove flying weather-vane listing has six images of a bird silhouette, arrow, cardinal assembly and pole, including an assembly montage and dimension graphic. Dedicated screws and mounting pieces belong to that host. Roofs, fence and landscape are not supplied goods, and setup, dimensions, durability and operating claims remain unverified.

Mara keeps the six-pair witness beside her chain-count note. The final noncontainment collection has no nested pair of selected cards, permits overlap and attains the maximum six. The twenty-four chains explain the bound and why equality requires pairs. Her fictional tabletop project can end with that complete mathematical record, leaving every actual product or purchasing question with the separate evidence it needs.

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