Check a nonempty box assignment count two ways

A nonempty box assignment count needs a rule for handling exclusions that overlap. In this original exercise, four distinct cards must each go into one of three labeled boxes, and every box must receive a card. There are 36 allowed assignments. Starting with all 81 assignments, subtracting three missing-box counts removes some assignments twice. Restoring those overlaps gives 81 minus 48 plus 3, or 36.

The cards and boxes are stipulated mathematical objects. Their labels identify them; no physical container capacity, actual stock or random procedure is assumed. A second calculation uses the only possible occupancy pattern, two cards in one box and one in each other box. That count also gives 36. The two calculations help check different possible errors in defining and recording the same outcomes.

Define what makes two assignments different

Call the cards A, B, C and D, and call the boxes L, M and R. Every card goes into exactly one box. Several cards may share a box. There is no order among cards inside a box and no stacking position to record. An assignment is completely specified by writing the destination of A, then B, then C, then D in a fixed record order.

For example, the record LLMR sends A and B to L, C to M and D to R. The record LMLR sends A and C to L, B to M and D to R. These are different assignments even though both have two cards in L. Distinct card identities matter. Reversing the written order of cards inside L would leave either assignment unchanged, because internal order is outside the definition.

Box identities matter too. LLMR differs from MMLR: the first puts the paired cards in L, while the second puts them in M. Permuting the box names generally changes an outcome. Keep this convention explicit throughout the worksheet. Dividing the final answer by six would impose an equivalence between renamed boxes that the exercise has not supplied.

Count every assignment before imposing the condition

Card A has three possible destinations. After one destination has been chosen for A, card B still has three, because boxes have no capacity limit in the exercise. The same applies to C and D. The multiplication 3 times 3 times 3 times 3 counts all 81 records of length four over the symbols L, M and R.

This total includes records such as LLLL, which leaves two boxes empty, and LLMR, which fills all three. It also includes records that use exactly two boxes, such as LLRR. The nonempty requirement has not yet been applied. Counting this larger, easily described collection gives a starting point from which to remove the forbidden assignments.

An exact enumeration can produce the 81 records by letting each of the four recorded positions independently run through L, M and R. That procedure describes an exhaustive finite list. It makes no statement about a physical person placing cards, and it does not assign equal likelihood to the records. Counting possibilities and prescribing a random sampling method are separate tasks.

Subtract the assignments missing each box

There are 16 assignments that leave L empty. Every card then chooses between M and R, giving 2 times 2 times 2 times 2. The same reasoning gives 16 assignments missing M and 16 missing R. Subtracting these three counts from 81 gives 33. This intermediate answer has removed every forbidden assignment, but some removals have been repeated.

Consider MMMM. It belongs to the assignments missing L and also to the assignments missing R. The subtraction therefore removes it twice, although it appeared only once in the original total. Its running contribution is one minus one minus one, or negative one. A forbidden assignment should contribute zero to the final count. The extra subtraction needs a correction.

By contrast, LLLM misses only R. It appears in exactly one missing-box count, so its contribution becomes one minus one, or zero. LLMR misses no box and is never subtracted, leaving its contribution at one. Following these individual records identifies the counting error directly: it occurs precisely among assignments that put all four cards in a single box.

Restore the pairwise overlaps once

An assignment missing both L and M must put every card in R. There is exactly one such record, RRRR. Missing L and R similarly forces MMMM, while missing M and R forces LLLL. Thus the three pairwise intersections each contain one assignment. Add these three counts back to the intermediate total of 33, obtaining 36.

Each one-box assignment is now counted correctly. For MMMM, the starting contribution one was reduced by two missing-box subtractions and then increased by the intersection for missing L and R. Its final contribution is one minus two plus one, or zero. The correction restores the over-subtracted contribution without admitting that record as a valid result.

There is no assignment missing all three boxes. Four cards must each have a destination, and none would be available. The triple intersection therefore contributes zero. The calculation ends at 81 - 3 times 16 + 3 times 1 - 0 = 36. No additional adjustment is needed for this particular set of premises.

Audit the contribution of every occupancy type

An assignment using all three boxes has no missing box. Its contribution is one from the starting total and zero from every correction, leaving one. An assignment using exactly two boxes has one missing box. It is subtracted once and occurs in no pairwise overlap, leaving zero. An assignment using one box has two missing boxes, so the calculation above leaves zero as well.

These three descriptions exhaust the possibilities because at least one box is used when four cards each have a destination. The correction consequently counts every allowed assignment once and every forbidden assignment zero times. This per-assignment audit explains the calculation independently of the final arithmetic. A matching total obtained from incorrectly combined cases would not supply that explanation.

The number 45 is the resulting count of forbidden assignments, since 81 minus 36 equals 45. Adding the three individual missing-box counts gives 48, which exceeds 45 by exactly the three repeated removals. A worksheet that simply writes 81 minus 48 would leave this discrepancy unresolved. Recording the named overlaps beside their corrections makes the source of the difference inspectable.

Count the occupancy pattern independently

Every valid assignment puts four cards into three nonempty boxes. Begin by allocating one card to each box; only one of the four cards remains beyond those three. Therefore exactly one box contains two cards and each other box contains one. There cannot be two paired boxes, because that would require at least five cards while keeping the third box nonempty.

Choose the two cards that will share a box. The possible pairs are AB, AC, AD, BC, BD and CD, giving six choices. Choose which of the three labeled boxes receives that pair. The two remaining cards must then occupy the two remaining boxes separately. They can do so in two ways, giving 6 times 3 times 2 = 36 assignments.

There is no repetition in this second count. A completed valid assignment identifies its unique two-card box, the unique pair inside it and the destinations of the two remaining cards. Those facts recover exactly the choices that generated the assignment. Conversely, any choices in this construction fill every box and place all four cards once. This establishes a direct check on both completeness and uniqueness.

Avoid changing the outcome while checking the answer

Recording only occupancy numbers would answer another question. The occupancy record two, one, one does not say which cards form the pair. Even specifying that L contains two cards leaves six possibilities for that pair and two ways to place the remaining cards. Counts of occupancy patterns and counts of assignments use different definitions of what distinguishes an outcome.

Internal order would change the exercise as well. If the paired cards were assigned a first and second position inside their box, every valid assignment would acquire two arrangements. Such positions have not been defined here. The factor of two already present in the occupancy count assigns the two singleton cards to different boxes; it does not order the pair inside its box.

The four-card derangement exercise assigns each label once to fixed positions and forbids fixed points. This exercise lets labels share a destination and requires every destination to be used. The allowed objects and the reasons for excluding an outcome differ.

The two-group selection exercise counts an unordered subset through mutually exclusive composition cases. Here the missing-box cases overlap. Their shared records need a correction before the remaining assignments can be counted.

Explain the alternating correction for larger exercises

Suppose there are n distinct cards and k labeled boxes, with n and k positive whole numbers. For any chosen set of j boxes to leave empty, each card has k minus j possible destinations. There are C(k, j) ways to choose that missing-box set, where C(k, j) counts its j-element subsets. The resulting signed sum is the sum over j from zero through k of (-1) to the j times C(k, j) times (k - j) to the n.

To verify that expression, fix one complete assignment and inspect its contribution. If it leaves no box empty, only the empty missing-box set applies, contributing one. If it leaves some boxes empty, choose one particular empty box. Pair every subset of its empty boxes that omits this chosen box with the subset obtained by adding it. Their sizes differ by one, so their signs are opposite and their contributions cancel.

That pairing leaves every forbidden assignment with total contribution zero. It proves the finite formula directly, without relying on a forecast or extrapolating a table. The assumption that n is positive also makes the final term with no available destinations zero. When n is smaller than k, no assignment can fill all boxes, and the same cancellation argument leaves a total of zero.

The saved exact control enumerates all 81 four-card records, retains the 36 that use every box and lists the 45 rejected records. It also checks 24 finite parameter pairs: n ranges from one through six and k from one through four. In each case, direct enumeration agrees with the signed sum. Those checks support the arithmetic; the cancellation argument explains why the expression works for all positive sizes.

Retain the limits of the native housing evidence

The FeedingTime Chalet Birdhouse listing shows a white architectural house with a peaked shingled roof, circular opening, balcony rails and a rear mounting piece against a tree. The tree and garden are staging. The image does not verify dimensions, material, ventilation, species suitability or successful nesting.

The Genuiua gourd-house listing has six views of red, green and tan bulbous housings with circular openings and hanging cords. A plant, apple and glass are staging. Material, coating, weather resistance and bird safety remain unverified. These houses supply no mathematical assignment premise.

For the nonempty box assignment count, keep the outcome definition beside the calculation. Four distinct cards, three labeled destinations and no empty destination give 36 assignments. The overlap correction counts each forbidden record zero times, while the unique two-one-one occupancy construction confirms the result through a separate enumeration of choices.

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