Mathematics Tutor Singapore | JC H2 Maths permutations and combinations questions are rarely hard just because a student cannot compute factorials. The larger difficulty is defining what makes two outcomes different, deciding whether order matters and respecting restrictions. One extra phrase such as ‘must be together’, ‘no repetition’ or ’round table’ changes the mathematical sample space.
Did you know that selecting three students out of six for a group gives twenty possible committees, but assigning gold, silver and bronze medals to those six gives one hundred and twenty possibilities? The same people are being selected, but the medals distinguish the positions and the committee does not.
This guide uses the 2026 Singapore-Cambridge H2 Mathematics 9758 Probability syllabus to explain counting principles, permutations, combinations, repeated objects, restricted linear and circular arrangements, committee cases and counting probabilities. All examples are original and include independent checks rather than formula-only answers.
The first decision: what is one complete outcome?
Before choosing nPr or nCr, say what one outcome means. If six pupils are available and a committee of three has no named roles, the outcome is one three-person collection. Selecting A,B,C is the same committee as C,A,B. By contrast, if the pupils fill three ranked positions, A,B,C differs from C,A,B.
A useful rule is to swap two chosen objects mentally. If the result counts as a new outcome under the story, order matters. If nothing about the outcome changes, the selection may be unordered. A restriction about adjacency, distinctness or labels can then further reduce the count.
- Define whether each object is distinguishable.
- Check whether order or roles matter.
- Mark whether repetition is permitted.
- Read whether positions are in a line, circle or labelled seats.
- Count only arrangements satisfying every restriction.
- Test that disjoint cases do not overlap.
- Verify using a smaller example, complement or alternative count.
Addition and multiplication principles
Use addition for mutually exclusive ways to complete a selection. If someone chooses exactly one of three morning activities or one of four afternoon activities, there are 3+4=7 possible choices, provided the categories are disjoint.
Use multiplication for choices made at successive stages. With five shirts and three trousers, there are 5×3=15 shirt–trouser outfits if all combinations are allowed. More complex counting often multiplies inside each distinct case, then adds across cases.
| Problem | Calculation | Count |
|---|---|---|
| Choose one talk from 3 morning or 4 afternoon | 3+4 | 7 |
| Pair a shirt from 5 and trousers from 3 | 5×3 | 15 |
| Assign two different roles from 8 pupils | 8×7 | 56 |
| Choose two pupils without roles from 8 | 8C2 | 28 |
Factorials, permutations and combinations
The factorial 5! represents 5×4×3×2×1=120, the number of ways to line up five distinct items. The convention 0!=1 is consistent with counting one empty arrangement. For n distinct items chosen r at a time without replacement, nPr=n!/(n−r)!.
For an unordered group of r selected from n distinct items, use nCr=n!/[r!(n−r)!]. This divides the ordered count by r! because the internal permutations of the same group are not new committee outcomes.
| Selection | Formula | Answer |
|---|---|---|
| Rank 3 from 6 | 6P3=6×5×4 | 120 |
| Select 3 from 6 | 6C3=6!/(3!3!) | 20 |
| Choose 5 from 8 | 8C5 | 56 |
| Select 2 role-free members from 8 | 8C2 | 28 |
A quick numerical check uses 6C3=20 and 6P3=120. Each three-person group can be assigned the three ranks in 3!=6 ways, and 20×6=120. The relationship explains why these formulas differ.
Worked example: gold, silver and bronze
Six distinct finalists compete for gold, silver and bronze. No person can take more than one prize. Gold has six choices, silver five after gold is assigned, and bronze four after that. Therefore 6P3=6×5×4=120.
Using 6C3=20 would count only the chosen set of medal winners, ignoring which winner receives which medal. I ask students to compare outcomes ABC and BAC. If the letter positions name different awards, these outcomes are distinct.
Worked example: five-person committee from eight
Eight volunteers are available and five are selected to a committee with no assigned offices. Their order on the list does not matter, so the number is 8C5=56. Equivalently, choosing five to serve determines the three who are not chosen: 8C3=56.
But if the same eight candidates are assigning a chairperson and secretary, the offices make order relevant: 8×7=56 assignments. That numerical equality to 8C5 is a coincidence; the underlying sample spaces are completely different.
Two named people adjacent in a straight line
Five distinct friends A,B,C,D,E line up, and A and B must be adjacent. Treat A and B temporarily as one block. We now have four objects: the block, C, D and E. Their order can be chosen in 4!=24 ways.
Inside the block, AB and BA are the two allowed orders. Therefore the total is 2×4!=48. The block method enforces adjacency without individually testing every arrangement.
| Stage | Ways | Reason |
|---|---|---|
| Arrange block plus three people | 4!=24 | Four line objects |
| Order A and B inside block | 2 | AB or BA |
| Total | 48 | Multiply independent choices |
Two people not adjacent: use the complement
There are 5!=120 unrestricted ways to line up A,B,C,D,E. Of these, forty-eight have A and B together. The rest have A and B separated, giving 120−48=72.
The restrictions ‘together’ and ‘not together’ are mutually exclusive and cover every line arrangement. Their counts 48 and 72 add to 120, providing a useful independent check. This is often more reliable than counting many separate non-adjacent cases.
Alternating arrangements of three and three
Suppose three distinct women and three distinct men line up in alternating categories. Their category pattern must be WMWMWM or MWMWMW. For either pattern, the women can be arranged in 3!=6 orders and the men in 3!=6 orders.
Total=2×3!×3!=2×6×6=72 line-ups. The two starting patterns are disjoint. The names of the people within each stated category are distinct, while only the category pattern is being fixed.
This is a purely mathematical problem using supplied categories. In real contexts one should not infer someone’s gender from their appearance or name.
No two women adjacent: the gap method
A different line-up contains four distinct men and three distinct women, and the three women may not sit or stand next to each other. Arrange the four men in 4!=24 ways. This creates five gaps: before the first man, between successive men and after the last.
Choose three distinct gaps from five: 5C3=10. Arrange the three women into the chosen gaps: 3!=6. The result is 24×10×6=1440.
| Decision | Calculation | Count |
|---|---|---|
| Arrange four men | 4! | 24 |
| Choose three among five gaps | 5C3 | 10 |
| Arrange three women | 3! | 6 |
| Total | 24×10×6 | 1440 |
The gap method depends on all selected women being separated. If the question allowed exactly one adjacent pair, we would need a different partition of cases. Teaching one method as universally applicable would be unsafe.
Arranging repeated identical letters
The word LEVEL has five letters: two Ls, two Es and one V. Counting all five labels as distinct gives 5!=120, but swapping the two identical Ls or two identical Es does not produce a visibly new word arrangement. Divide by 2! and 2! to correct these duplicates.
Thus LEVEL has 5!/(2!2!)=30 distinct linear arrangements. BANANA contains three As, two Ns and one B, giving 6!/(3!2!)=60.
| Letters | Repeated counts | Distinct permutations |
|---|---|---|
| LEVEL | L twice, E twice | 30 |
| BANANA | A three, N twice | 60 |
| MOM | M twice | 3 |
| BOOK | O twice | 12 |
If a student multiplies by the repeated factorials instead of dividing, their result may exceed the total for all-distinct letters. That is a warning sign: making objects indistinguishable cannot create more visible arrangements.
Circular permutations: fix an anchor
Six distinct friends sit around an unlabelled round table. Arrangements differing only by a rotation of the entire seating order are treated as identical; mirror reflections are treated as different unless the question specifies otherwise.
Fix one named person as an anchor, then arrange the remaining five people clockwise. There are 5!=120 circular arrangements. In general this convention gives (n−1)! for n distinct people.
| Situation for six people | Count | What is distinguishable? |
|---|---|---|
| Standing in a line | 6!=720 | Left and right positions |
| Round table with rotations identical | 5!=120 | Relative circular order |
| Six numbered seats | 6!=720 | Each labelled seat |
| Round table with fixed A–B adjacency | 2×4!=48 | Relative circular order and block orientation |
For the adjacent A–B circular case, combine them into one block, leaving five circular objects. These have (5−1)!=4!=24 relative orders, and the block has two orientations, yielding 48.
The number with A and B not adjacent is 120−48=72. The numerical result matches a previous five-person line case by coincidence, not a universal identity.
Four-digit numbers from 0,1,2,3,4
Make four-digit integers using digits 0,1,2,3,4 with no repetition. The first digit cannot be zero, giving four choices. Then four digits remain for the second place, three for the third and two for the last.
There are 4×4×3×2=96 such integers. The unadjusted 5P4=120 would count leading-zero strings, which do not represent four-digit numbers even though they could be legitimate four-character codes.
Even four-digit integers require separate cases
For an even number, the final digit is zero, two or four. When the final digit is zero, the first has four choices and the middle positions have three then two, so count 4×3×2=24.
When the final digit is two or four, choose it in two ways. The leading digit has three nonzero choices excluding that final digit. The middle two positions have three and then two choices. Count 2×3×3×2=36.
The total is 24+36=60. Trying to use one uniform product for all three possible last digits often double-counts or allows a leading zero.
| Final digit case | Calculation | Count |
|---|---|---|
| Zero | 4×3×2 | 24 |
| Two or four | 2×3×3×2 | 36 |
| All legal even numbers | 24+36 | 60 |
A committee requiring exactly two women
A group has five distinct women and four distinct men. A four-person committee must contain exactly two women and two men, with no named roles. Choose two from five women in 5C2=10 ways and two from four men in 4C2=6 ways.
Multiply the independent subgroup selections: 10×6=60 committees. This counts membership, not seating or ranked positions. If a chairperson were subsequently assigned, that would create an additional distinct selection stage.
At least two women: enumerate or use a complement
The same committee of four must now contain at least two women. Exactly two women contributes 5C2×4C2=60. Exactly three women contributes 5C3×4C1=40. Exactly four women contributes 5C4×4C0=5.
The disjoint cases total 60+40+5=105. For a complement check, all committees number 9C4=126. Exclude zero women: 4C4=1. Exclude exactly one woman: 5C1×4C3=20. The remaining count is 126−1−20=105.
| Allowed female count | Ways |
|---|---|
| Exactly 2 | 60 |
| Exactly 3 | 40 |
| Exactly 4 | 5 |
| At least 2 total | 105 |
| Complement verification | 126−1−20=105 |
Use combinations in a probability model
A bag holds five red and three blue distinguishable balls, and two balls are drawn uniformly at random without replacement. If only the unordered pair matters, the total number of equally likely pairs is 8C2=28.
Both red: choose two from the five reds, 5C2=10, giving probability 10/28=5/14. At least one red: exclude the blue-blue pairs, 3C2=3, giving (28−3)/28=25/28.
| Event | Favourable pairs | Probability |
|---|---|---|
| Both red | 5C2=10 | 5/14 |
| Both blue | 3C2=3 | 3/28 |
| At least one red | 28−3=25 | 25/28 |
We could also count ordered draws, provided both numerator and denominator use ordered counts. Mixing an unordered favourable count with an ordered total would produce the wrong probability.
Common exam mistakes and their first causes
| Observed mistake | Likely cause | Next check |
|---|---|---|
| Uses combinations for ranked awards | Order ignored | Does swapping medal roles change the outcome? |
| Uses permutations for a committee | Internal orders counted as new | Does swapping names change membership? |
| Counts repeated letters with n! alone | Indistinguishable swaps included | Which letters repeat? |
| Uses 6! for round-table seating | Rotations counted separately | Can one person be fixed? |
| Divides by 2 for circular reflections without instruction | Model not read | Are mirror seatings stated equivalent? |
| Leading zero allowed in a four-digit integer | Place-value definition ignored | Is it still four-digit? |
| Counts only exactly two for at least two | Valid cases omitted | What other female counts qualify? |
| Counts one red-blue order only | Alternative outcomes omitted | Does blue-red also qualify? |
| Treats no adjacency as simple adjacent block | Restriction reversed | Is a complement easier? |
| Adds cases that overlap | Double-counting | Can one arrangement belong to both? |
A pupil can enter nCr correctly yet count the wrong kind of outcome. In tuition, I ask for a brief sentence defining distinguishability and then one small example or complement check before the final numerical answer.
Eighteen original H2 counting questions
1. Outfits
Five shirts and three trousers yield 15 outfits.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
2. One talk
Three morning or four afternoon talks, choose exactly one: 7.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
3. Rank three from six
6P3=120.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
4. Choose three from six
6C3=20.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
5. Committee of five from eight
8C5=56.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
6. Two named offices from eight
8P2=56.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
7. Two role-free members from eight
8C2=28.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
8. Adjacent friends in line
Five people with A,B together: 2×4!=48.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
9. Non-adjacent friends in line
Five people with A,B apart: 5!−48=72.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
10. Alternation in a line
Three distinct women and three distinct men alternating: 72.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
11. Gap method
Four distinct men and three distinct women, no two women adjacent: 1440.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
12. LEVEL
5!/(2!2!)=30.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
13. BANANA
6!/(3!2!)=60.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
14. Six-person unlabelled round table
(6−1)!=120, rotations identified.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
15. Circular two-person block
Six around a table with A,B adjacent: 48.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
16. Four-digit nonzero-leading numbers
Digits 0–4 without repeats: 96.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
17. Even four-digit numbers
Same digits and constraints: 60.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
18. Four-member committee, at least two women
Five women and four men: 105.
Check what is distinguishable, whether the selected objects can repeat and whether the condition applies to a line, circle, role or label. Where possible, verify by a complement or a simpler test case.
A JC Maths revision pathway
| Session | Focus | Independent test |
|---|---|---|
| 1 | Addition versus multiplication | Defines a complete outcome |
| 2 | nPr and nCr | Explains order |
| 3 | Ranked awards and role-free groups | Selects formula |
| 4 | Adjacent and separated people | Uses block and complement |
| 5 | Alternating and gap restrictions | Counts valid positions |
| 6 | Repeated identical letters | Divides factorials |
| 7 | Circular unlabelled arrangements | Fixes an anchor |
| 8 | Leading zero and last-digit restrictions | Separates cases |
| 9 | Exactly and at least committee counts | Adds disjoint cases |
| 10 | Probability with combinations | Consistent sample space |
An effective tutor should use the student’s first incorrect counting expression to identify whether the problem lies in order, identical objects, rotational equivalence or interpreting a restriction. A changed problem after feedback then tests transfer.
Frequently asked permutations and combinations questions
How do I know when to use nPr?
Use ordered selections where swapping two assigned positions creates a different outcome.
How do I know when to use nCr?
Use unordered selections where only which objects are chosen matters.
Why is nCr smaller than nPr?
A group of r people appears in r! different orders in an ordered count.
What is 0!?
It equals one by convention, corresponding to one empty arrangement.
Why do repeated letters require division?
Swapping identical letters does not produce a visibly distinct arrangement.
Why is round-table seating often (n−1)!?
When rotations are treated as identical, fixing one person’s position removes the rotational copies.
Are reflections of round-table arrangements identical?
Not automatically. The question must specify whether mirror images count as the same.
How does the gap method prevent adjacency?
It places selected people into separate gaps created after the other people are arranged.
What does at least two mean in a committee problem?
It includes exactly two and every higher feasible count.
Are circles and repeated arrangements part of 2026 H2 9758?
Yes. SEAB Section 6.1 includes arrangements in a line or circle with repetition and restriction.
Can better counting guarantee an A-Level A?
No. It is one important Probability and Statistics skill, but examination outcomes depend on much more.
2026 SEAB syllabus and related Mathematics reading
The 2026 H2 Mathematics 9758 syllabus Section 6.1 includes counting principles, permutation and combination, linear and circular arrangements with repetition and restrictions, probability rules and methods using counting. Always follow the student’s actual examination-year document.
Continue through the Mathematics Learning Hub, Secondary 2 Venn Diagrams, Secondary 4 Probability Trees, and JC H2 Discrete Random Variables.
The original Secondary 1 Mathematics Tutor Clementi small-group reference remains unchanged. For an individual Mathematics consultation, contact eduKate Singapore with one unrestricted counting problem and a failed restricted arrangement.