BTT Mathematics / Primary Mathematics Learning Hub / Systematic Listing
Systematic listing means organising possible answers so that every valid possibility appears once and nothing invalid is counted. It is useful when a question asks how many arrangements, combinations, routes, numbers or mixtures are possible. The final count matters, but so does the reason the list is complete.
Suppose two shirts, red and blue, can each be worn with three pairs of shorts, black, white and green. Fix the red shirt and pair it with all three shorts. Then fix the blue shirt and repeat. The six outfits are easy to verify because every shirt choice has been handled and every shorts choice appears under it.
Now change the question to choosing two colours from four, where choosing red then blue is the same selection as choosing blue then red. An ordered list of choices would count that pair twice. The method must begin by deciding what counts as a different answer, not by writing possibilities as quickly as possible.
This guide develops lists, tables and simple branching records; distinguishes order from selection; handles repetition and leading zero; and checks restricted routes, coin totals and incomplete data. It is a worked companion to the broader word-problem guide, with a specific emphasis on completeness.
The MOE Primary Mathematics syllabus, pages 11 and 13–14, discusses mathematical representations, reasoning and problem-solving strategies. The organising rules, examples and 24 practice questions here are original. Later counting puzzles are enrichment selected by prerequisite knowledge, not a claim about the content of a particular examination.
What counts as different? · Organise the list · Digit and order problems · Totals and mixtures · Routes and restrictions · Prove completeness · 24 questions · Worked solutions · Teaching and transfer
1. Define one valid answer before counting many answers
Every counting problem needs a clear description of the objects being counted. An outfit consists of one shirt and one pair of shorts. A two-digit number has a non-zero tens digit and a units digit. A committee consists of a selected group of people, not an order in which their names are written.
Write one candidate and test it against the wording. Does it contain the right number of objects? Does it use allowed choices? Can a choice repeat? Does changing the order produce a genuinely different result? These decisions define the collection you are about to count.
Order matters in some questions
Placing A before B in a two-person queue is different from placing B before A. The arrangements AB and BA are two distinct queues. Similarly, the numbers twenty-four and forty-two are different because the digits occupy different place values.
However, selecting A and B for an unranked pair does not become a new selection when the names are written as BA. Both descriptions select the same two people. A list designed for ordered arrangements needs adjustment before it can count unordered selections.
Repetition must be stated or established
With letter tiles A, B and C, making a two-position pattern may allow AA if tiles can be reused. If only one physical copy of each tile is available, AA is impossible. The difference changes the number of choices available for the second position.
Do not assume repetition simply because a symbol can be written twice. The problem defines whether choices can be reused. When the condition is missing, explain the alternative interpretations rather than silently selecting one.
Identical objects do not create new selections by swapping places
If two plain red counters are indistinguishable, exchanging them does not create a visually or mathematically different colour arrangement. If the counters are labelled R1 and R2 and the labels matter, their exchange may create a different arrangement.
The question’s definition of identity controls the count. This is why a count can change when objects receive labels even though the same physical number of counters remains on the table.
Choose a consistent representation
For an unordered pair of numbers, write the smaller number first. For a colour selection, keep a fixed alphabetical order. For a route, use R for one step right and U for one step up. A consistent record makes it possible to compare candidates and identify duplicates.
The representation is not the answer itself. It is a way of recording the answer unambiguously. Explain what every letter or position represents before using a short code in the working.
2. Hold one choice fixed while exhausting the others
The simplest systematic list fixes the first choice, lists every allowed continuation and then moves to the next first choice. This creates groups that do not overlap and together cover all possibilities.
Example A: Outfits
With red or blue shirts and black, white or green shorts, list red-black, red-white and red-green. Then list blue-black, blue-white and blue-green. There are six outfits. Every outfit has one of the two shirts, and each shirt has been paired with each of the three shorts exactly once.
The multiplication 2 × 3 = 6 summarises that organisation. It works because every first choice has the same three valid second choices. The list explains why multiplication is appropriate rather than leaving it as a rule about seeing two groups of objects.
Example B: A restriction changes the rows
Suppose the red shirt cannot be worn with green shorts. The red group now contains two outfits, while the blue group still contains three. Add the group sizes: two plus three gives five. Alternatively, begin with all six and remove the single forbidden outfit.
Multiplying two by three without accounting for the restriction would count an invalid result. Restrictions must be reflected in the branches or removed from an already well-defined complete list. Do not apply both approaches and remove the same invalid case twice.
When a table is useful
A table with shirts as rows and shorts as columns displays each pair at one intersection. A crossed-out cell records a forbidden pair. The table is particularly useful when restrictions depend on both choices, because each decision has a visible place.
A table does not need to be elaborate. Its job is to give every candidate one location. If the learner writes the same pair in different cells, the row and column meanings are not yet stable.
When a branching record is useful
For three successive choices, a simple tree or indented list can keep the order clear. Begin with one branch for each first choice. From each branch, record the second choices still allowed. Continue with the third choices. One complete path from start to finish represents one complete arrangement.
Count completed paths, not all branch points. An unfinished branch represents a partial decision. Adding partial and complete paths together would mix objects of different sizes and overcount the requested arrangements.
A boundary prevents an endless search
A question about coin combinations totalling a fixed amount has an upper limit on how many large coins can be used. A question about whole-number rectangle sides with a fixed area has a finite set of factor pairs. Write those limits before listing.
Without a stopping rule, a child may stop because the page is full rather than because every case has been handled. A systematic method explains both how to begin and why it is safe to stop.
3. Digit problems test place value as well as counting
Example C: Two-digit numbers without repetition
Use the digits two, five and eight to make two-digit numbers without repeating a digit. Fix two in the tens place: twenty-five and twenty-eight. Fix five: fifty-two and fifty-eight. Fix eight: eighty-two and eighty-five. The complete list contains six numbers.
Each of the three tens choices has two remaining units choices, so 3 × 2 = 6. The list and product agree. The number twenty-two is excluded not because repeated digits are generally forbidden, but because this particular question forbids repetition.
Example D: A zero cannot begin an ordinary two-digit number
Use zero, four and nine without repetition to make two-digit numbers. The tens digit may be four or nine, but not zero. With four, the numbers are forty and forty-nine. With nine, they are ninety and ninety-four. There are four valid numbers.
Writing 04 does not make a two-digit number in ordinary number notation; it represents four. A two-position label may allow 04, but that would be a different kind of object. Read whether the task asks for a number or a fixed-length pattern.
Example E: A two-position label that permits zero and repetition
Make two-position labels using zero, one and two, allowing repeated symbols and zero at the beginning. Fix the first symbol at zero to get 00, 01 and 02. Then obtain 10, 11 and 12; then 20, 21 and 22. There are nine labels.
This is not inconsistent with the previous example. The rules changed: labels retain their two positions even when the first is zero, and symbols may repeat. Counting begins with the definition of a valid result.
Example F: An even-number restriction
Use digits one, three, four and six without repetition to make two-digit even numbers. The units digit must be four or six. With four in the units place, the tens digit can be one, three or six: fourteen, thirty-four and sixty-four. With six last, obtain sixteen, thirty-six and forty-six.
There are six numbers. Here fixing the last digit first is more efficient than fixing the first digit, because the parity restriction applies directly to the last position. A systematic list need not always begin from the left; it should begin with a useful constraint.
Check bounds as well as digits
A number may use allowed digits and still fall outside a required interval. For instance, a question asking for numbers below fifty cannot include sixty-four from the previous list. Apply every condition before counting the candidate.
For several restrictions, write a short checklist: allowed digits, repetition rule, leading digit, required parity and numerical range. The aim is to avoid inventing or forgetting a condition while the list grows.
4. List mixtures and pairs without losing the total
Example G: Unordered pairs with a fixed sum
Find pairs of positive whole numbers that sum to twelve, treating reversed pairs as the same. Start with the smaller number one: the partner is eleven. Continue with two and ten, three and nine, four and eight, five and seven, then six and six.
Stop there. A first number of seven would require a partner of five, repeating the pair already written as five and seven. The rule “first number no larger than the second” prevents duplication and supplies the stopping point. Six pairs are possible.
Example H: Coins with a fixed value
Find ways to make one hundred forty cents using only twenty-cent and fifty-cent coins. Coins of the same denomination are not distinguished, and either denomination may have count zero. Consider the possible fifty-cent counts: zero, one or two. Three would already exceed the total.
With zero fifty-cent coins, the remaining one hundred forty cents requires seven twenty-cent coins. With one fifty-cent coin, ninety cents remains and cannot be made from twenty-cent coins exactly. With two fifty-cent coins, forty cents remains, requiring two twenty-cent coins. There are two valid denomination mixtures.
The rejected middle row belongs in the reasoning because it shows that the case was checked rather than forgotten. The upper limit on large coins proves no later cases are possible. This is a complete search, not merely the discovery of two successful examples.
Example I: Three packet sizes
Five packets contain either one, three or five counters each, with seventeen counters altogether. Let the number of five-counter packets increase systematically. With none, even five three-counter packets contain only fifteen, so that case fails. With one large packet, the remaining four packets must contain twelve counters, forcing four medium packets.
With two large packets, the remaining three packets must contain seven counters, giving two medium and one small. With three large packets, the remaining two counters come from two small packets. Four large packets would already contain twenty counters and exceed the target. These three mixtures exhaust the cases.
This is a useful extension beyond the two-type assumption method. One type is fixed first, then the remaining count and total constrain the two other types. Every partial choice must be updated before the next calculation.
Example J: Rectangles with whole-number dimensions
A rectangle has area thirty square units and positive whole-number side lengths. Treat turning the rectangle as the same pair of dimensions. The factor pairs are one by thirty, two by fifteen, three by ten and five by six. There are four possibilities.
A rectangle with sides six and five is already counted as five by six. A rectangle with sides four and seven has area twenty-eight, not thirty, so it is invalid. Listing factor pairs provides both a construction method and a product check.
Area and perimeter impose different conditions
For a rectangle with perimeter twenty-four units, the two adjacent sides sum to twelve. Its possibilities come from positive unordered sum pairs, not factor pairs of twenty-four. Confusing these constraints can create a beautifully organised list of answers to the wrong problem.
Before choosing a list structure, translate the given geometry into the equation or sentence it actually imposes: side times side for area, or twice the sum of adjacent sides for perimeter.
5. Routes and arrangements need clearly named moves
Example K: Shortest paths on a small grid
Move from the bottom-left corner of a two-by-two grid to the top-right corner, using only one-unit steps right or up. Every route contains two R moves and two U moves. The six move strings are RRUU, RURU, RUUR, URRU, URUR and UURR.
The strings record complete routes. Two strings with the same letters in different orders can pass through different intermediate points and therefore represent different routes. A string containing a left move or three right moves would not satisfy the stated rules.
Example L: A forbidden point
Now forbid the central point reached after one right and one up move. Four of the six strings pass through it: RURU, RUUR, URRU and URUR. The two remaining routes are RRUU and UURR.
Trace each rejected route until it first reaches the forbidden point. A route that touches the point is invalid even if it leaves immediately afterwards. The restriction concerns every intermediate position, not merely the final destination.
Example M: A fixed place in an arrangement
Three labelled cards A, B and C are arranged in a row, with B in the middle. Once B is fixed, only A and C remain for the two ends. The two rows are ABC and CBA.
This is simpler than generating all six rows and crossing out four. Both approaches are valid, but fixing the strongest restriction early reduces unnecessary work. The shorter method should still explain why both end orders have been included.
Example N: A simple assignment puzzle
Ana, Ben and Chen each choose one different fruit from apple, banana and orange. Ben chooses banana. Ana does not choose apple. Since all choices differ, Ana cannot choose banana either and must choose orange. Chen therefore chooses apple.
This is elimination rather than trial by guess. Every exclusion has a reason, and the final assignment satisfies all conditions. If the question removed the condition about Ana, two assignments would remain. A missing condition can turn a unique answer into several possible answers.
Do not infer equal likelihood from a list
A list of possible outcomes says what can happen under the rules. It does not automatically say how likely each outcome is. A mechanism may favour some choices, and different underlying arrangements can produce the same recorded result.
In this guide, counting tasks do not require probability unless explicitly stated. A learner should not turn “three possible totals” into “each has probability one-third” without a justified model of how outcomes are generated.
6. Three checks make a list trustworthy
Validity: every listed result satisfies the rules. No duplication: no result has been counted more than once. Completeness: every allowed result has a place in the list. A correct count needs all three, even when the numbers are small.
Explain where any missing answer would have to go
For two-digit numbers, any valid number must begin with one of the permitted tens digits. If every allowed units digit has been listed beneath each tens digit, there is no unexamined place for another valid number. That is the completeness argument.
For coin totals, any valid mixture must have one of the large-coin counts within the established limit. If every such count has been tested, all mixtures have been considered. A statement about the structure of the cases is stronger than “I could not think of any more.”
A second count can reveal a mistake
Three different letters in two ordered positions without repetition give three first choices and two second choices, or six arrangements. If the written list contains only five, the discrepancy shows that at least one case is missing or the multiplication model has been misread.
The check should not be used blindly. If restrictions make different branches have different numbers of continuations, multiplying two fixed counts may be invalid. Return to the actual branch sizes and add them.
Distinguish one answer, all answers and a count
“Give one arrangement,” “list all arrangements” and “how many arrangements are there?” are different tasks. A single valid example answers only the first. A count without supporting organisation may be insufficient to explain the third.
A complete list can support both the second and third tasks, but the final sentence should still report what was asked. Some problems ask for the largest or smallest valid candidate rather than the number of candidates. Use the completed list to make that final selection.
7. Practice: 24 questions
State whether order matters, whether repetition is allowed and how you will know when to stop. Keep worked solutions covered. In questions involving coin or packet counts, counts may be zero unless another condition says otherwise. All objects with different printed labels are distinguishable.
Questions 1–8: Define and organise possibilities
1. Two shirts, red and blue, can each be paired with black, white or green shorts. List every outfit.
2. Use digits 1, 3 and 5 to form two-digit numbers without repetition. List them all.
3. Use digits 0, 2 and 7 to form ordinary two-digit numbers without repetition. List them all.
4. Make two-position labels using 0, 1 and 2. Repetition and an initial zero are allowed. List every label.
5. Place two different tiles from A, B and C in a row. Order matters. List all rows.
6. Choose two tiles from A, B, C and D, without repetition. Order does not matter. List all selections.
7. List positive whole-number pairs that sum to ten. Reversed pairs count as the same pair.
8. List ordered pairs of positive whole numbers that sum to five. Here reversed unequal pairs count separately.
Questions 9–16: Totals and restrictions
9. Find every way to make 100 cents using only 20-cent and 50-cent coins. Count denomination mixtures, not coin order.
10. Packets contain either three or five counters. Find every packet-count mixture totalling nineteen counters.
11. Five packets contain one, three or five counters each, totalling seventeen. List all mixtures by counts of the three packet types.
12. List positive whole-number side pairs for rectangles of area twenty-four square units. Rotations count as the same pair.
13. List positive whole-number side pairs for rectangles of perimeter twenty units. Rotations count as the same pair.
14. Use digits 1, 2, 4 and 7 without repetition to make two-digit even numbers. List every valid number.
15. Arrange A, B and C in a row with B in the middle. List all arrangements.
16. Arrange A, B and C in a row, with A not in the first position. List all arrangements.
Questions 17–24: Routes, logic and completeness
17. A route contains exactly three unit moves right and two unit moves up. List every possible order of these moves using R and U.
18. From the bottom-left to top-right of a two-by-two grid, use only right and up unit moves and do not pass through the centre. List all move strings.
19. Two labelled dice, first and second, each show a number from one to six. List the ordered face pairs whose sum is six. No probability calculation is required.
20. Choose two different letters from A, B, C, D and E, with A included. Order does not matter. List all selections.
21. Use only digits 0, 1, 2 and 3, with repetition allowed, to make two-digit even numbers from ten to thirty inclusive. List all valid numbers.
22. Ana, Ben and Chen each choose a different fruit from apple, banana and orange. Ben chooses banana, and Ana does not choose apple. Find each person’s fruit.
23. Use digits 1, 2, 3 and 4 without repetition to make two-digit numbers whose tens digit exceeds the units digit. List them all.
24. A learner lists 12, 14, 21, 24 and 41 as all two-digit numbers made from 1, 2 and 4 without repetition. Identify what is missing and explain how to prove the corrected list is complete.
8. Worked solutions and completeness checks
Solutions 1–8
1. Six outfits. Red-black, red-white, red-green, blue-black, blue-white and blue-green. The two shirt cases are separate, and each contains all three shorts choices. Every outfit has exactly one of those shirts, so these cases cover the whole set.
2. 13, 15, 31, 35, 51, 53. Fix each of the three possible tens digits in turn. Two different digits remain for the units position each time, giving six numbers. The reversed pairs are different numbers and should remain in the list.
3. 20, 27, 70, 72. The tens digit cannot be zero. With two first, use zero or seven; with seven first, use zero or two. These two groups exhaust the allowed non-zero leading digits, producing four ordinary two-digit numbers.
4. 00, 01, 02, 10, 11, 12, 20, 21, 22. Each of the three first symbols permits all three second symbols. There are nine labels. Initial zero is permitted because the question defines fixed-position labels rather than ordinary two-digit numbers.
5. AB, AC, BA, BC, CA, CB. Each first tile leaves two different second tiles, giving six rows. AB and BA are distinct because order matters. AA, BB and CC fail the requirement that the two tiles differ.
6. AB, AC, AD, BC, BD, CD. Write each selection in alphabetical order. Three include A with a later letter, two begin with B and a later letter, and one begins with C. This counts each of the six unordered pairs once.
7. (1,9), (2,8), (3,7), (4,6), (5,5). Always write the smaller value first. Stop when the values meet, since later pairs reverse earlier ones. Each pair sums to ten, and five pairs are possible.
8. (1,4), (2,3), (3,2), (4,1). The first positive value can be one through four. Its partner is five minus that value. Since order matters here, the two reversals are retained rather than removed as duplicates.
Solutions 9–16
9. Five 20-cent coins; or two 50-cent coins. The 50-cent count can only be zero, one or two. Those cases leave one hundred, fifty or zero cents for 20-cent coins. Only the first and third divide exactly into twenties. Hence there are two denomination mixtures.
10. Three three-counter packets and two five-counter packets. The five-counter count can be zero, one, two or three. Subtracting their contributions from nineteen leaves nineteen, fourteen, nine or four. Only nine is divisible by three. The unique mixture therefore contains three small and two large packets.
11. Counts (0,4,1), (1,2,2), (2,0,3). The entries count one-, three- and five-counter packets respectively. Each triple sums to five packets and contributes seventeen counters. Testing the five-counter count from zero to three gives exactly these cases; four would already exceed seventeen counters.
12. 1×24, 2×12, 3×8, 4×6. These are the positive factor pairs of twenty-four with the smaller side written first. Reversed side pairs are rotations of the same rectangle and are not counted again. There are four possibilities.
13. 1×9, 2×8, 3×7, 4×6, 5×5. Half the perimeter is ten, so adjacent side lengths must sum to ten. The five unordered positive sum pairs give every whole-number rectangle. Multiplying to twenty would impose the wrong condition.
14. 12, 14, 24, 42, 72, 74. An even units digit must be two or four. Each choice permits three different tens digits. Listing by units digit gives 12, 42, 72 and 14, 24, 74; sorting yields the displayed complete list of six.
15. ABC and CBA. Fix B in the middle. A and C can occupy the two ends in exactly two orders. No arrangement with another middle letter satisfies the restriction.
16. BAC, BCA, CAB, CBA. If B is first, the remaining letters can be AC or CA. If C is first, they can be AB or BA. These are all possibilities because the first letter cannot be A.
Solutions 17–24
17. Ten routes. RRRUU, RRURU, RRUUR, RURRU, RURUR, RUURR, URRRU, URRUR, URURR and UURRR. Each contains three R and two U moves. Grouping by the location of the first U gives groups of four, three, two and one routes, totalling ten.
18. RRUU and UURR. Starting with one R and one U in either order reaches the forbidden centre. The route must therefore make its first two moves in the same direction. The remaining two moves then go in the other direction, giving exactly two valid routes.
19. (1,5), (2,4), (3,3), (4,2), (5,1). Fix the first die from one to six and find the second value needed for a sum of six. A first value of six would require zero, which is not allowed. Five ordered pairs remain.
20. AB, AC, AD, AE. A is already selected. The second letter can be any one of the other four. Writing BA instead of AB would not create a new unordered selection.
21. 10, 12, 20, 22, 30. The units digit must be zero or two. A tens digit of one or two allows both. A tens digit of three allows only zero because thirty-two exceeds the upper bound. Zero cannot be the tens digit of an ordinary two-digit number.
22. Ana: orange; Ben: banana; Chen: apple. Ben’s choice removes banana from the others. Ana is also forbidden apple, leaving orange. The remaining fruit is apple for Chen. Every step follows a stated exclusion, so the assignment is uniquely determined.
23. 21, 31, 32, 41, 42, 43. A tens digit of one has no smaller allowed units digit. Two permits one; three permits one and two; four permits one, two and three. The case counts are zero, one, two and three, totalling six.
24. 42 is missing. Group the corrected list by tens digit: 12 and 14; 21 and 24; 41 and 42. Each of the three leading digits has exactly two permitted different partners. Six numbers are required, and the grouped list includes all six without duplication.
9. Teach the organisation, not only the final count
When a learner misses an answer, ask which group of the list should contain it. This is more useful than supplying the missing candidate immediately. The aim is to repair the organising rule so that a similar omission becomes easier to prevent in a new question.
When the learner lists randomly
Choose a small example and require one fixed first choice until every continuation is exhausted. A child who wants to jump from a red-shirt outfit to a blue-shirt outfit can mark the unfinished red group and return to complete it. The written organisation should show which parts remain open.
Once the child can manage two choices, introduce a third. Keep the candidate set small enough to inspect. Increasing the number of possibilities without improving the organising method mostly increases the opportunity for omissions.
When the learner double-counts reversed pairs
Place two labelled counters in a selection box and ask whether swapping their written order changes which counters were chosen. Then place the same counters in two marked positions and ask whether swapping changes the arrangement. The physical comparison separates an unranked selection from an ordered placement.
Return to notation afterwards. Adopt a fixed alphabetical or numerical order for recording unordered selections. This is a bookkeeping rule that ensures one representation per selection, not a claim that the selected objects themselves have an order.
When one condition disappears during the list
Write the conditions beside the working and test each candidate against them. A digit may be allowed but in the wrong position; a sum may be correct but the object count wrong; a route may finish correctly after passing through a forbidden point.
Use a short code such as a tick beside each completed condition during teaching, then remove that support gradually. The goal is accurate reasoning with a manageable record, not permanently lengthy checking for every simple candidate.
Use a deliberately incomplete list
Offer a list that contains one missing result and one duplicate. Ask the learner to reorganise it rather than merely correct the total. This turns completeness into an observable task. A learner may notice the duplicate quickly but still need a systematic structure to locate the missing case.
Do not infer a broad ability level from one successful repair. Use another example with a changed representation, such as moving from letter arrangements to digit numbers where zero has a special restriction.
A short teaching sequence
Begin by defining a valid outcome. Make a small list together. Ask why no case is missing. Change one rule, such as allowing repetition, and rebuild the list. Finally ask the learner to select a list, table or tree for a fresh question without being told which representation to use.
Record the reasoning that changed: “I now fix the larger coin count and check every remainder” is more useful than “coin problems practised.” Distinguish a supported list from one completed independently, and revisit the relationship with new values before extending the size of the search.
Connect a complete list to a justified conclusion
A complete list can establish that exactly one solution exists, that several solutions exist or that none exist. All three are legitimate mathematical outcomes. The conclusion should reflect the full checked set rather than an expectation that every school-style question must produce one neat number.
This is the wider value of systematic listing: it makes the search inspectable. A reader can see which choices were admitted, which were rejected and why the process has finished.
Continue through the problem-solving collection
For a two-type total with a constant replacement difference, use The Assumption Method and Guess-and-Check. For reconstructing earlier stages, use Working Backwards and Reverse Processes. For reducing the search through a preserved total, difference or parity, use Invariants, Before-and-After and Unchanged Quantities.
Use Factors, Multiples and Divisibility when rectangle or exact-group listings need a number-structure repair. Return to the BTT Primary Mathematics Learning Hub for all learning routes.
Original examples and 24 original questions with separate worked solutions. All rules and restrictions apply only as stated in each problem. Curriculum reference accessed 6 September 2026. No official examination coverage or outcome is promised.