Build a two-group selection count without double-counting

A two-group selection count can be checked by counting the same choices in two ways. In this original card exercise, one group contains three distinct labels and another contains four. Select three distinct cards from the combined seven, without recording their order. There are thirty-five possible selections. Sorting them by how many cards came from the first group gives four case totals: four, eighteen, twelve and one.

The cards and labels are invented inputs for a mathematical exercise. Each label identifies one card, and no label occurs in both groups. The task does not describe a stockroom, a customer preference or a random draw. Its purpose is to explain why the case counts multiply within each case, why those cases add, and how to check that a selection has not entered the total twice.

Define a selection before counting it

Call the left group A, B and C. Call the right group P, Q, R and S. A selection is a set of exactly three of these seven labels. Once a card has been selected, it cannot be selected again. The written strings ABP and PBA therefore describe the same selection. Alphabetical notation is only a convenient way to keep a record consistent; it does not create positions within the selected set.

For example, ABP contains two left-group cards and one right-group card. APQ contains one from the left and two from the right. PQR contains none from the left, while ABC contains all three left cards. These are all allowed. There is no requirement to take a card from each group, and membership in a group does not make its cards interchangeable. A and B remain distinct choices.

These definitions decide which calculation applies. If the task instead recorded a first, second and third card, reversing two cards would usually produce another outcome. If replacement were allowed, a label might occur repeatedly. Neither condition applies here. Keeping the unordered, without-replacement definition beside the count prevents a later calculation from quietly changing the object being counted.

Obtain the direct total from seven labels

Use C(n, k) to mean the number of k-element subsets of a set with n distinct elements. Thus C(7, 3) counts the selections in the exercise directly. One way to calculate it is to first count ordered lists of three different labels. There are seven choices for the first entry, six remaining choices for the second and five for the third, giving 7 times 6 times 5, or 210 lists.

Each unordered three-card selection appears in exactly six of those lists. Its three distinct members can occupy the first position in three ways, then the second in two ways, with the last member forced. Dividing 210 by 6 gives 35. The division is justified by a uniform repetition count: every selection has three distinct members and therefore exactly the same number of orderings.

This direct total is a useful check on the case calculation below. It uses the whole seven-label set without reference to the division into groups. A disagreement between the two answers would require revisiting the definitions or the arithmetic. Agreement alone would not prove that each case was constructed correctly, so the case argument will also explain how every selection receives a unique home.

Separate four possible left-group counts

Let j be the number of selected cards from the left group. A selection has three cards altogether, so j can be zero, one, two or three. Once j is fixed, exactly 3 minus j selected cards must come from the right group. These four values exhaust the possibilities because a count cannot be negative or exceed the total selection size.

The cases cannot overlap. A particular selection has one definite left-group count, even when its labels are written in a different order. ABP has count two every time it is recorded. It cannot also belong to the count-one case. This is the reason the four case totals can eventually be added: they describe separate portions of the same collection of allowed selections.

Within a fixed case, choose the left subset and the right subset independently. Each allowed left subset can be joined to every allowed right subset of the required size. The groups have no labels in common, so joining them creates exactly three distinct labels. That gives C(3, j) times C(4, 3 minus j) selections in the case. The multiplication counts pairs of subsets, not possible orders of individual cards.

Calculate the zero and one cases

For j equal to zero, take the empty subset from the left and a three-card subset from the right. There is exactly one empty subset, so C(3, 0) equals one. The four right triples are PQR, PQS, PRS and QRS. The case total is therefore 1 times 4, or 4. Taking nothing from a group supplies one possible subset in this case, rather than eliminating the case.

For j equal to one, the left subset is A, B or C, giving three choices. The right pairs are PQ, PR, PS, QR, QS and RS, giving six. Any one left label can be joined to any one of these pairs. The total is 3 times 6, or 18. For instance, APQ and BPQ are different selections because their left members differ.

A compact check writes three rows, one for each left label, and six columns, one for each right pair. Each cell names the union of its row label and column pair. No two cells name the same selection: the left member identifies its row and the right pair identifies its column. This small grid makes the multiplication visible without introducing an extra ordering of the three chosen cards.

Calculate the two and three cases

For j equal to two, the left pairs are AB, AC and BC. The right singleton can be P, Q, R or S. There are three choices followed by four choices, giving 12 selections. ABP is one cell in this three-by-four arrangement. ACP occupies another cell because its left pair differs, while ABQ differs through its right singleton. Each change produces a different set of labels.

For j equal to three, choose ABC from the left and the empty subset from the right. Each choice is forced. The total is C(3, 3) times C(4, 0), or 1 times 1, giving one selection. The empty right subset contributes no card, while still supplying the single partner needed for the pair-of-subsets description.

Adding the four cases gives 4 + 18 + 12 + 1 = 35. Their different sizes reflect the stipulated group sizes and selection size. A case with more left cards need not have more possibilities. Here the count-one case is largest because it combines three left singletons with six right pairs. The arithmetic follows the available subsets in each group.

Check the split and join in both directions

Start with any allowed selection and split it according to its original groups. Its left part is the intersection with A, B, C; its right part is the intersection with P, Q, R, S. This produces exactly one ordered pair of subsets, with left membership determining j. A label cannot migrate between those parts because the original groups are fixed and disjoint.

Conversely, start with a left subset of size j and a right subset of size 3 minus j. Their union is an allowed selection of size three. Splitting that union again recovers exactly the two subsets used to form it. Joining the split parts of an original selection also recovers that original selection. These inverse operations explain both completeness and the absence of repeated counting.

The pair has a left role and a right role, but that does not impose an order on the selected cards. The grouping information is already attached to each label. There is consequently no additional C(3, j) factor for choosing imagined positions among three slots. Such a factor would answer an ordered-placement question that the selection exercise never asked.

Extend the same reasoning to other finite sizes

Let the disjoint groups instead have n and m distinct members, and choose an unordered subset of size r. For any feasible left count j, there are C(n, j) choices on the left and C(m, r minus j) on the right. Their product counts that case. Summing over all feasible j counts every r-element subset of the combined group once, by the same split-and-join argument.

The resulting equality is the sum over j of C(n, j)C(m, r minus j) = C(n + m, r). Define C(a, b) as zero when b is negative or greater than a, for nonnegative integer a. With that convention, impossible group-size choices contribute zero. If r exceeds n + m, both the direct count and all case totals are zero because too few distinct members exist.

The boundary r equal to zero also fits. The only selection is the empty set. It splits into one empty left subset and one empty right subset, giving a single pair. An empty original group is handled in the same way. These checks help distinguish a genuine zero count from the one choice associated with selecting no members of an available group.

Use exact controls and retain the model limits

The saved finite control lists all thirty-five three-card selections from the seven labels and partitions them by their actual left count. The resulting bucket sizes are 4, 18, 12 and 1, and joining the recorded parts recovers every original selection. A second exact control checks 252 finite parameter combinations: n and m each range from zero through five, while r ranges from zero through n plus m plus one.

Those controls test the arithmetic and the boundary convention in a finite range. The split-and-join argument supplies the general reasoning beyond that range. No probability follows merely from having thirty-five choices. To assign equal probability to each one, an exercise would need an additional selection procedure or an explicit probability assumption. This count assigns neither.

The four-card derangement exercise counts assignments to fixed positions with no fixed points through cycle types. Here each outcome is an unordered subset, and its group composition supplies the cases; there are no matched positions or cycle restrictions.

The noncontainment collection exercise bounds a whole family of subsets under a containment restriction. This exercise counts individual three-card selections from seven labels, with no restriction comparing one selected set against another.

Keep the native product evidence separate

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For a two-group selection count, record the distinct labels, whether order matters and whether replacement is allowed before doing arithmetic. With the stipulated three-plus-four groups and three-card unordered selection, the direct total is 35. The unique group split explains the four products and their sum, while the reverse join confirms that no allowed selection is missing or repeated.

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