A list of possible answers can look convincing and still be wrong. Every entry may satisfy the question, yet one valid case is missing. Alternatively, the list may contain all the possibilities but count the same outcome twice under different names. In PSLE Mathematics, checking a list therefore requires more than checking the arithmetic inside each row.
This guide develops a specific examination habit: prove that your list is complete without repeating a case. The word prove here means giving a clear reason that your search covered every allowed possibility and counted each outcome once. You do not need advanced notation or a formula for arrangements. You need a sensible order, a clear definition of one case and a stopping point justified by the question.
Use the existing Primary 6 guide to systematic listing and case organisation to learn the underlying method. This companion workshop concentrates on auditing the list after you have started: finding missing branches, removing duplicates, checking endpoints and deciding when the search is genuinely finished. All questions below are original practice examples.
Three checks are needed, not one
The first check is validity. Does each entry obey the conditions? If the question requires a three-digit even number without repeated digits, every listed number must have three digits, end in an even digit and avoid repetition. An entry that fails any condition must be removed or corrected.
The second check is completeness. Have all allowed entries been considered? A list containing 124 and 132 may contain only valid entries, but it is not a complete list of the three-digit even numbers that can be formed from 1, 2, 3 and 4 without repetition. Correct examples do not establish that nothing is missing.
The third check is uniqueness. Does every row represent a different outcome? If a question asks for a pair of pupils, Ana with Ben and Ben with Ana may be the same pair. If it asks for a captain and an assistant, exchanging their roles creates a different outcome. The question decides what counts as different. You must make that decision before counting rows.
Define one case in a full sentence
Before making a list, finish the sentence: “One case means…” For a number problem, one case may mean one three-digit number. For a money problem, one case may mean a particular number of each coin type. For a rectangle problem, one case may mean one pair of side lengths, with a turned rectangle counted as the same shape.
This definition protects the count from changing halfway through. Imagine beginning a coin problem by counting combinations of denominations, then adding another case because the same coins can be placed in a different order on the table. You have changed the meaning of a case without changing the question. The extra row is not a new solution to the original task.
Labels help. Write “number of 50-cent coins” rather than just “50”, or “shorter side” rather than “side”. A row should identify a complete outcome clearly enough that another person can tell whether it is new. If two rows have different handwriting but the same labelled quantities, they are duplicates.
Choose an anchor that cannot be skipped
An anchor is the feature you hold fixed while listing the remaining possibilities. It might be the hundreds digit, the number of large coins, the shorter side of a rectangle or the first member of a pair under a chosen alphabetical rule. Choose an anchor with a small, clearly bounded set of values.
Work through those values in order. If the anchor can be 0, 1, 2, 3 or 4, show all five stages, even when a stage produces no valid outcome. Writing “none” for an impossible stage can be useful because it distinguishes a rejected branch from a forgotten branch.
Do not rely on inspiration to produce the next case. Random listing makes it hard to know whether you have searched a region already. A consistent anchor changes the question from “What else can I think of?” to “What remains possible while this value is fixed?” That is a much more checkable task under time pressure.
Worked example 1: an even-number list with a hidden gap
Use the digits 1, 2, 3 and 4 to form three-digit even numbers smaller than 400. No digit may be repeated. A learner writes 124, 132, 214, 234, 312 and 342, then reports six possibilities. Every listed number is valid. The mistake is not in the individual entries; it is in the coverage of the search.
Use the hundreds digit as the anchor. Because the number is smaller than 400, the hundreds digit can be 1, 2 or 3. Because the number is even, the units digit must be 2 or 4. Once those two positions are chosen, the tens digit must come from the unused digits. These conditions create a short, organised search.
For hundreds digit 1, ending in 2 gives 132 and 142. Ending in 4 gives 124 and 134. That branch contains four numbers. For hundreds digit 2, the number cannot also end in 2, so it must end in 4. The unused tens digits give 214 and 234, making two numbers. For hundreds digit 3, ending in 2 gives 312 and 342; ending in 4 gives 314 and 324. That branch contains four.
The complete count is 4 + 2 + 4 = 10. In increasing order, the numbers are 124, 132, 134, 142, 214, 234, 312, 314, 324 and 342. The completeness explanation is not “I cannot think of more”. It is that all permitted hundreds digits and all permitted even endings have been checked, with every remaining tens digit considered.
Use a second organisation to check the first
The same number list can be checked by fixing the units digit first. Numbers ending in 2 can begin with 1 or 3, because 2 cannot be repeated and a hundreds digit of 4 would make the number too large. Each permitted hundreds digit leaves two choices for the tens digit, giving four numbers ending in 2.
Numbers ending in 4 can begin with 1, 2 or 3. Each of those choices leaves two available tens digits, giving six numbers ending in 4. The second organisation therefore gives 4 + 6 = 10. It reaches the same count through a different grouping.
This is a useful independent check because the missing cases in a random list may become obvious under another grouping. However, two matching counts are not enough if both methods accidentally use the same wrong condition. Before comparing totals, confirm that both searches use the same meaning of three-digit number, even ending, no repetition and smaller than 400. A well-organised search of the wrong problem is still wrong.
Worked example 2: coin combinations are not coin arrangements
How many combinations of 20-cent and 50-cent coins make exactly $2? Either type may be absent. Define one case as the count of 20-cent coins together with the count of 50-cent coins. The order in which the coins are placed on a table does not create another combination.
Fix the number of 50-cent coins. It can be 0, 1, 2, 3 or 4. Five would already exceed $2, so there is a justified stopping point. With no 50-cent coins, ten 20-cent coins make $2. With one 50-cent coin, $1.50 remains, which cannot be made using whole 20-cent coins. With two 50-cent coins, $1 remains, giving five 20-cent coins.
With three 50-cent coins, 50 cents remains, which again cannot be made using whole 20-cent coins. With four 50-cent coins, nothing remains, so zero 20-cent coins are needed. The valid pairs, written as counts of 20-cent coins and 50-cent coins, are (10, 0), (5, 2) and (0, 4). There are three combinations.
Check each pair by rebuilding the amount. Then check that every possible number of 50-cent coins from zero to four was considered. Finally, check that moving coins around would not change either count. These are three different checks. The arithmetic verifies each row; the bounded anchor verifies coverage; the definition of a combination prevents duplicate counting.
Zero can be a valid endpoint
Now change the coin question: at least one coin of each type must be used. The same search structure still works, but the allowed endpoints change. The combinations (10, 0) and (0, 4) are no longer valid because one type is absent. Only (5, 2) remains.
A common error is excluding zero automatically in the original question because it feels as though both mentioned types must appear. Another error is allowing zero automatically after the question explicitly requires both types. Neither habit is safe. Read the condition and decide whether zero is permitted for each quantity.
This endpoint check transfers to many questions. Can a pupil receive no tokens? Can a box be empty? Must both colours appear? Can the shorter side of a rectangle be zero? The answers depend on the situation. In a rectangle with positive side lengths, zero is not a valid side. In a coin combination where a type may be absent, zero is a valid count. Completeness includes the correct endpoints, not every imaginable endpoint.
Worked example 3: rectangles and rotated duplicates
A rectangle has a perimeter of 24 cm. Both side lengths are whole numbers of centimetres. How many different pairs of side lengths are possible, counting a rectangle and its rotation as the same shape? The perimeter relationship gives length + width = 12.
Choose the shorter side as the anchor. It must be at least 1 cm. It cannot be greater than 6 cm because then both sides would exceed 6 cm and their sum would exceed 12 cm. The cases are therefore 1 and 11, 2 and 10, 3 and 9, 4 and 8, 5 and 7, and 6 and 6. There are six possible pairs, including the square.
Why do we stop at 6 and 6? The next pair, 7 and 5, describes the same side lengths as 5 and 7. Continuing through 8 and 4, 9 and 3 and so on would count rotations again. The rule “shorter side first” gives every unordered pair one consistent written form.
Add a second condition: the area must be at least 32 square centimetres. The areas are 11, 20, 27, 32, 35 and 36 square centimetres. Only 4 by 8, 5 by 7 and 6 by 6 remain, so there are three qualifying rectangles. The list is complete because every permitted shorter side was checked, and it has no repeated shape because each pair appears with the shorter side first.
Do not divide by two unless every case has a partner
A tempting shortcut is to list length and width in both orders and divide the total number of rows by two. That works only if every outcome has been counted exactly twice. The square is a warning: 6 and 6 does not produce a different reversed row. It pairs with itself.
If you list all positive whole-number ordered pairs adding to 12, there are eleven rows: from (1, 11) through (11, 1). Dividing eleven by two does not produce the correct count of shapes. Ten non-square rows form five reversal pairs, while (6, 6) is one additional shape. The result is six, not a rounded version of five and a half.
The safer primary-school method is to prevent duplicates rather than repair them with an unexplained division. Choose a consistent ordering rule before listing. Use smaller side first, alphabetical pair order or another rule that matches the question. Then explain why every allowed outcome has exactly one representation under that rule. That reasoning is more reliable than applying the same numerical shortcut to every listing task.
Worked example 4: a group and a pair of roles are different
Five fictional pupils, Ana, Ben, Chen, Devi and Eli, are available for a two-person team. Exactly one of Ana and Ben must be on the team. The roles are identical. List the possible teams.
Use two disjoint branches: Ana is included and Ben is excluded; or Ben is included and Ana is excluded. With Ana, the second person can be Chen, Devi or Eli. With Ben, the second person can also be Chen, Devi or Eli. The six teams are Ana–Chen, Ana–Devi, Ana–Eli, Ben–Chen, Ben–Devi and Ben–Eli.
Ana–Ben is invalid because it includes both of the named pupils. Chen–Devi is invalid because it includes neither. Chen–Ana is not an additional team because it contains the same two people as Ana–Chen. The branch rule checks the exact-one condition, while the definition of a team prevents order from creating duplicates.
Now change the task to choosing a captain and an assistant from the same eligible pairs. The roles are different. For every eligible two-person team, either person can be captain, so each team creates two role assignments. There are twelve assignments. The mathematics changed because the definition of an outcome changed, not because the names changed. Always establish whether position or role makes a new case before counting.
Worked example 5: codes are not ordinary numbers
Make a three-character code using 0, 1 and 2. Repetition is allowed. The code must contain exactly one zero and end in an even digit. A code may begin with zero. How many codes are possible?
Separate the search by the final character. If the code ends in 0, that is already its one zero. Each of the first two positions can be 1 or 2. The four codes are 110, 120, 210 and 220. If the code ends in 2, exactly one of the first two positions must contain zero. When the first position is zero, the second can be 1 or 2, giving 012 and 022. When the second position is zero, the first can be 1 or 2, giving 102 and 202.
The total is eight codes. Each belongs to exactly one ending branch, and the position of the zero prevents overlap within the branch ending in 2. A learner who discards 012 and 022 because a number cannot begin with zero has imported the wrong definition: the question asks for codes and explicitly permits a leading zero.
Check the conditions separately. Repetition is allowed, so 220 and 022 are valid. Exactly one zero is required, so 002 and 200 are invalid. The code must end in an even digit, so 021 is invalid. A short code problem can therefore contain several different traps, and a complete list must obey all of them simultaneously.
Worked example 6: two conditions can leave very few cases
A practice shop sells small packs costing $2, medium packs costing $3 and large packs costing $4. Buy exactly three packs for a total of $10. More than one pack of a size may be bought. List the possible numbers of each size.
Use the number of large packs as the anchor. It can be zero, one or two; three large packs would cost $12 and exceed the target. With no large packs, even three medium packs cost only $9, so there is no solution in that branch. With one large pack, the remaining two packs must cost $6. They must be two medium packs.
With two large packs, $8 has been spent and one pack remains. That pack must be small, costing $2. The two valid combinations are zero small, two medium and one large; or one small, zero medium and two large. Both use three packs and cost $10.
Notice the completeness reasoning. The maximum number of large packs is justified by cost, and every count from zero to that maximum is checked. Within each branch, both the remaining cost and remaining number of packs are used. A learner who checks only cost might include five small packs; a learner who checks only pack count might include three medium packs. Both would solve only part of the problem.
Worked example 7: a new condition changes the valid list, not its history
Choose two different activity sessions from lengths of 30, 45, 60 and 75 minutes. Their total activity time must be no more than 105 minutes. The order of the sessions does not matter. Using the shorter session first, the six possible pairs have totals of 75, 90, 105, 105, 120 and 135 minutes.
The qualifying pairs are 30 with 45, 30 with 60, 30 with 75, and 45 with 60. There are four. Equality is allowed because no more than 105 minutes includes exactly 105. Each pair appears once because the shorter session is written first.
Now the question adds a 15-minute break between the two sessions and keeps the overall limit at 105 minutes. The activity time must therefore be no more than 90 minutes. Only 30 with 45 and 30 with 60 remain. You can filter the existing complete list rather than invent an entirely new search.
This is a useful examination move. When a later part adds a restriction, keep the original cases visible and mark which survive. Do not delete the earlier reasoning so thoroughly that you can no longer explain the first part. A complete earlier list becomes a resource for the next part, provided that the meaning of a case has not changed.
Worked example 8: overlapping conditions create repeated cases
Consider the whole numbers from 1 to 30. List those divisible by 3, by 5 or by both. A learner counts ten multiples of 3 and six multiples of 5, then adds the counts to obtain sixteen. The two component lists are individually complete, but the combined count is wrong because some numbers belong to both lists.
The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27 and 30. The multiples of 5 are 5, 10, 15, 20, 25 and 30. The numbers 15 and 30 appear in both. Counting them again does not create new outcomes. The correct total is fourteen distinct numbers.
One way to organise the answer is to count all ten multiples of 3, then add only the multiples of 5 that were not already counted: 5, 10, 20 and 25. The branches now have no overlap. Another way is to combine the two lists in increasing order and write each number once. Both methods work because they respect the definition of one case: one number in the stated interval.
This example reveals a different source of duplication from reversed pairs. Nothing was turned around. Instead, the same outcome entered through two conditions. Whenever you add the counts from separate groups, ask whether a case could qualify for more than one group. If it can, either use non-overlapping groups or account for the repeated cases explicitly. Do not assume that different descriptions always identify different outcomes.
A correct total can hide an incorrect list
Return to the three-digit example. Suppose a learner writes 124, 132, 134, 132, 214, 234, 312, 314, 324 and 342. There are ten written entries, which matches the correct total. However, 132 appears twice and 142 is missing. A duplicate and an omission have cancelled in the count without cancelling in the mathematics.
This is why checking only the number of rows is insufficient when the task asks for the list itself. Sort the entries or group them by a fixed feature. In the hundreds-digit-1 branch, the repeated 132 and absent 142 become easier to notice. The branch should contain two possibilities ending in 2 and two ending in 4, with different tens digits in each pair.
For review, ask the learner to explain how the original mistake occurred. Did the eye return to a row already written? Did the learner count tick marks rather than unique outcomes? Did an unlabelled branch make the same case seem new? The repair should change the recording system, not merely replace one item after someone supplies the missing answer.
A relaxed condition may require reopening the search
Adding a restriction often lets you filter a complete existing list. Removing a restriction is different. Suppose the three-digit even-number problem no longer requires the number to be smaller than 400. The original list of ten is still valid, but it is no longer complete because a hundreds digit of 4 is now permitted.
Open the new branch. A number beginning with 4 cannot end in 4 because repetition is forbidden, so it must end in 2. The middle digit can be 1 or 3, giving 412 and 432. The enlarged list contains twelve numbers. Merely checking the original ten again would never reveal those two new cases if the old bound remained in your thinking.
The practical question is whether a changed condition narrows or widens the allowed set. A tighter limit may remove cases. A looser limit may add branches. A different definition, such as changing a team into two named roles, may change what counts as an outcome altogether. Identify which change occurred before reusing an earlier answer.
How to explain that your search is complete
A good completeness explanation has a beginning, a progression and an end. State the smallest possible anchor, explain how you moved through all its permitted values and justify why the next value is impossible or a duplicate. You do not need a long essay, but the logic should be visible in the table or working.
For the coin problem: the number of 50-cent coins runs from zero to four because five exceeds $2. For the rectangle problem: the shorter side runs from one to six because larger choices reverse an earlier pair. For the number problem: every permitted hundreds digit is combined with every permitted even ending and every unused tens digit.
These explanations establish more than confidence. They identify why an omitted case cannot be hiding outside the searched branches. The statement “I checked carefully” describes your effort. The statement “all three possible hundreds digits were covered” describes the mathematical coverage. Prefer the second kind of explanation when you need to justify a list.
When a listing method should be abandoned
Systematic listing is not the best method for every question. If an anchor has hundreds of values and there is a direct relationship that finds the unknown, a long list may waste time. Begin by asking how many cases your plan will require. A small, bounded search can be efficient; an uncontrolled search is a warning to reconsider the representation.
Look for totals, differences, divisibility or an unchanged quantity that can reduce the search. In the pack example, exactly three packs is a strong limit. Without that condition, more combinations would need consideration. In a coin problem, choosing the larger denomination as the anchor usually produces fewer branches than starting with the smaller denomination.
Do not abandon a good list merely because it contains one impossible branch. An explicit impossible branch is useful evidence of coverage. Change method when the search itself is too large or poorly organised, not when a single case fails. A failed candidate can be progress if it removes one defined part of the possibility space.
Practice set: make the coverage visible
Question one: use 2, 5 and 8 to form two-digit numbers greater than 50 without repeating a digit. List them and explain why none are missing. Do not treat 52 and 25 as the same number; their positions have different values.
Question two: make exactly $3 using 20-cent and 50-cent coins. Either type may be absent. List the combinations as counts of each coin type. Then find how many combinations remain when at least one coin of each type is required.
Question three: list whole-number side-length pairs for rectangles with perimeter 20 cm, counting rotations as the same shape and including squares. Then select those with area at least 24 square centimetres. Explain the largest shorter side you need to consider.
Question four: choose a two-person team from Ana, Ben, Chen and Devi, with Ana required to be included. Count teams, not roles. Then change the question to a captain and an assistant using the same eligible pairs. Explain why the count changes.
Question five: form two-character codes from 0, 1 and 2. Repetition and leading zero are allowed. The code must contain no 1 and end in an even digit. List the codes. State which condition becomes redundant after the allowed characters have been restricted to 0 and 2.
Question six: in the pack problem with costs $2, $3 and $4, require exactly one pack of each size. Can the total be exactly $10? Explain without listing imaginary extra cases. A valid answer can be that no case exists when the conditions are inconsistent.
Practice answers and the reason each search stops
For question one, the numbers are 52, 58, 82 and 85. The tens digit can only be 5 or 8 because the number must exceed 50. With each permitted tens digit, either unused digit can occupy the units place. There are two branches with two entries each.
For question two, the pairs of counts of 20-cent coins and 50-cent coins are (15, 0), (10, 2), (5, 4) and (0, 6). The 50-cent count runs from zero to six, and odd counts leave an amount that cannot be made in whole 20-cent coins. Requiring both types removes the two endpoint combinations, leaving two.
For question three, the side-length pairs are (1, 9), (2, 8), (3, 7), (4, 6) and (5, 5). The shorter side need not exceed five. The qualifying areas are 24 and 25 square centimetres, so the pairs are (4, 6) and (5, 5).
For question four, the teams are Ana–Ben, Ana–Chen and Ana–Devi. There are three. Assigning captain and assistant creates two role orders for each team, giving six. For question five, the codes are 00, 02, 20 and 22. All allowed last digits are already even, so the even-ending condition adds no further restriction once 1 has been excluded.
For question six, one of each pack costs $2 + $3 + $4 = $9, not $10. There is no valid combination under those exact conditions. The absence of a solution follows from the fixed selection, not from failing to think of enough possibilities.
From a supported list to independent examination work
Begin practice with the anchor supplied. For example, ask the learner to use the number of 50-cent coins or the shorter rectangle side. Once the learner can complete and explain the search, remove the anchor prompt and ask which feature would make the shortest reliable list.
On a later day, change the surface context. A coin problem can become a pack problem; a team problem can become a choice of two activities. Ask the learner to explain whether order matters and which endpoints are allowed. These decisions should transfer even when the original numbers and nouns have disappeared.
Finally, mix listing questions with questions better solved another way. Independent performance includes choosing not to list when a direct relationship is more efficient. Review the method choice as well as the final count. A correct answer reached by a very long uncontrolled search is not as dependable under time pressure as a short, justified method.
Guidance for parents and tutors
When checking a child’s list, do not immediately point to the missing entry. Ask which branch should contain it. If the child omitted 142, for example, return to hundreds digit 1 and units digit 2. The missing tens digit then becomes a structural gap rather than one more isolated answer to memorise.
Separate the three feedback questions: Is this row valid? Is every allowed branch covered? Is any outcome represented twice? A learner can succeed at one and fail at another. Praising a neat list without checking its coverage can hide the actual problem.
Ask for a stopping reason. The learner should know why one more large coin exceeds the amount, why a larger shorter side repeats a rectangle or why all permitted positions have been covered. That reason is part of the method, not an optional sentence added after the teacher says the list is correct.
Frequently asked questions
Must I write every possible case in full?
Not always. A clearly organised table, a justified count within each branch or a direct relationship may be enough for the task. During learning, writing the cases helps reveal missing and duplicated outcomes. During timed work, choose the shortest method whose coverage you can still explain.
Is a matching second count proof that the answer is right?
It is useful evidence, especially when the second count groups cases differently. But both methods must use the correct conditions. Two methods can agree while both wrongly exclude zero or treat unordered pairs as ordered outcomes.
When does order matter?
Order matters when changing positions or roles changes the outcome the question asks about. The number 25 differs from 52. A captain–assistant assignment differs when the roles are exchanged. A two-person team with identical roles usually does not become a new team when the names are written in reverse order.
What should I do with an impossible branch?
Mark it as impossible and give the short reason during practice. Do not count it as an outcome. Keeping the rejected branch visible can demonstrate that it was considered rather than skipped. In a compact examination solution, the bound or arithmetic may make that rejection clear without a long explanation.
Can I stop after finding one answer?
Only if the question asks for one valid example and your example satisfies every condition. If it asks how many, all possibilities, the greatest value or the least value, finding one candidate does not normally finish the job. You need coverage or another argument that establishes the requested result.
Do I need advanced counting formulas?
No. The examples in this workshop use ordinary arithmetic, organised cases and clear definitions. Understanding the list is more important than applying a formula whose conditions you cannot explain. Use methods appropriate to the question and to the Mathematics you have been taught.
Official reference and the next learning route
The 2026 PSLE Mathematics syllabus includes interpreting information, mathematical reasoning and selecting suitable problem-solving strategies among its assessment objectives. This workshop is an original practice route for making a case-based solution checkable; it is not a claim that a particular question type will appear.
Return to the Primary 6 Mathematics Learning Hub for the wider subject route and the PSLE Learning Guide for the series. The final audit is: every row is valid, every permitted branch is covered and every outcome is counted once. A list is finished when those reasons are clear, not merely when no further example comes to mind.
