Unit 14

Ordering Deduction

About this unit

Place people, objects, or events in the correct order — or determine what must, could, or cannot be in a specific position — using a set of positional clues. Questions range from simple 4-person linear arrangements to complex 8-person circular seating, relay handovers, and multi-criteria ranking puzzles.

What types of questions will you face?

  • 1Linear ordering: 4-6 people in a row with "left of", "right of", and "not at end" constraints
  • 2Circular seating: 6-8 people around a table with "opposite", "adjacent", and relative-direction constraints
  • 3Relay/handover chain: trace an object passing through a sequence of handovers at specific times
  • 4Multi-criteria ranking: rank people by two criteria simultaneously (time AND score) using clues
  • 5Partial ordering: given 3 facts about relative positions, determine what else MUST or CANNOT be true
  • 6"Valid arrangement" questions: given 4 possible orderings, identify the one that satisfies all rules

Skills you will build

  • Tracking multiple positional constraints (before/after, adjacent, not adjacent) simultaneously
  • Building up valid arrangements step-by-step from the most constrained element first
  • Applying "opposite in a circle" and "adjacent" rules for circular arrangements
  • Tracing relay handover chains forward in time
  • Determining what must be true vs. what could be true when the arrangement isn't fully fixed
  • Using elimination to discard invalid arrangements quickly

By the end of this unit, you will be able to

  • Solve any 4-8 person linear or circular arrangement puzzle
  • Trace relay chains across time to find who holds an item at any point
  • Determine fixed positions and free positions in a partially constrained arrangement
  • Answer "must be true / cannot be true" ordering questions reliably under time pressure

Difficulty profile

The hardest unit by average score (3.82). Simple 3-4 element orderings with one chain are Easy; circular arrangements with 6-8 people and multiple constraint types, relay chains, and multi-criteria rankings are Very Difficult. Mastery here significantly boosts your overall TS score.

Exam tip: Ordering Deduction

Start with the person/item with the most constraints — they are easiest to place. Then use each placed element to constrain the remaining ones. For circular arrangements, fix one person's seat first (the one with a "directly opposite" clue) then fill in from there.

Sample Questions

Lesson 1 of 7Ordering DeductionIntroductory

When two clues both say someone is to the right of two other people, that person must be the rightmost of the three. Draw a simple left-to-right number line, place the person the clues point to on the far right, and see which answer statement is always true.

Two-clue three-person ordering questions (find who must be true) appear in the very easy band of OC TS. They are near-certain marks for students who draw a quick diagram rather than reasoning in their head.

The examiner checks whether students can combine two “A is right of B” constraints and correctly deduce that A must be the rightmost person — and recognise that the relative order of the other two remains unknown.

Three people stand in a row. Two clues each say Person A is to the right of one of the others. Students must find which of four statements — including “A is in the middle” and “A is furthest right” — must be true.

Best approach: Write positions 1 (left) to 3 (right). Place Person A at position 3 since both clues require A to be right of everyone else. The other two can swap positions 1 and 2. Test each answer option against both possible arrangements.

Question

Archie, Beau and Carter are standing in a row.

If Archie is to the right of Beau and Carter is to the left of Archie, which one of the following statements must be true?

  1. ABeau is furthest to the right.
  2. BCarter is furthest to the left.
  3. CArchie is in the middle.
  4. DArchie is furthest to the right.

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Lesson 2 of 7Ordering DeductionEasy

This is a great starting point for ordering problems: three clues, five teams, one winner to find. The secret is to always start with the clue that gives you a fixed position — once you anchor Yellow at 3rd, the other positions fall into place very quickly.

Three-clue chain ordering questions appear regularly in the easy range of OC TS and reward students who work through the clues in the right order rather than trying to hold everything in their head at once.

The examiner wants to see whether you can build a position chain from relative clues — "after Red but before Yellow" compresses three positions into one locked sequence. Students who try to guess from the options rather than building the chain often confuse 1st and 2nd.

Five teams or people compete in a race or ordering task. One clue fixes a position (e.g. "Yellow finished third"). A chain clue ties two others around that anchor (e.g. "Green after Red but before Yellow"). A final clue places the remaining two relative to the anchor. You must identify who came first.

Best approach: Always start with the fixed-position clue — it is your anchor. Then read the chain clue and ask: "If Yellow is 3rd, where must Red and Green go?" Place them, then use the final clue to slot in the last two. Write down your chain as you go — do not try to track five positions in your head.

Question

There are five school teams: Red, Yellow, Green, Orange and Blue. A member of each team competes in the 100m race. You don't see the race but a friend tells you the following facts:

  • Yellow finished third.
  • Green finished after Red but before Yellow.
  • Blue beat Orange but finished after Yellow.

Who won the race?

  1. ARed
  2. BGreen
  3. COrange
  4. DBlue

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Lesson 3 of 7Ordering DeductionEasy

Peter has a book for Fiona but cannot see her today. He gives it to Zhang at 09:00. Meanwhile Sam→Tim at 10:00, Tim→Ursula at 11:00, Ursula→Fiona at 12:00. The right-hand chain is complete — but Zhang is isolated. The missing link is a meeting that connects Zhang into the chain before the chain moves on without the book.

Short 4–5 person relay questions with one obvious missing link appear regularly in the easier half of OC TS chain questions. The trap options always involve the right people but wrong timing (too early, too late, or the person doesn't have the object yet). Checking timing order is as important as checking who meets whom.

The examiner tests whether students can identify a gap in a relay chain (Zhang has the book but is disconnected from the Tim→Ursula→Fiona chain) and select the extra meeting that correctly bridges the gap with valid timing. Trap D (Zhang→Ursula at 12:30) is designed to catch students who ignore the timing — Ursula already met Fiona at 12:00.

An object (book, card, package) must travel from Person A to Person Z. Several meetings are scheduled but one link in the chain is absent. One extra meeting is needed to bridge the gap. Trap options include: meetings with wrong timing (too late or before the object arrives), meetings involving people outside the main chain, and meetings between the right people but where neither person has the object yet.

Best approach: Draw the chain with times. Label who holds the object and when. Find the gap: who has the object but has no scheduled meeting with anyone on the path to the destination? Then find the option that fills this exact gap and check the timing — the meeting must happen after the gap person receives the object AND before the next person in the chain needs to pass it on.

Question

Peter needs to return a book to Fiona, but he will not see her today. However, he knows that the following meetings will happen:

Peter will meet Zhang at 09:00. Sam will meet Tim at 10:00. Tim will meet Ursula at 11:00. Ursula will meet Fiona at 12:00.

Which one of the following additional meetings would enable Fiona to receive the book today?

  1. AFiona will meet Sam at 09:30.
  2. BFiona will meet Tim at 11:30.
  3. CZhang will meet Tim at 10:30.
  4. DZhang will meet Ursula at 12:30.

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Lesson 4 of 7Ordering DeductionIntermediate

Relay Logic questions are one of the most practically engaging types in Ordering Deduction — an object must travel from one person to another through a chain of timed handovers, and your job is to find where the chain breaks and what single extra meeting would repair it. The approach is always the same: trace forward from where the object starts.

Chain of handovers questions appear regularly in the medium difficulty range of OC TS and reward students who think in chains rather than individually about each meeting. The most common mistake is picking a meeting that involves the right people but at a time before the object could have reached them.

The examiner wants to confirm that you can map a relay chain, identify exactly which link is missing, and verify that the proposed fix happens at the right time — after the object arrives with the sender AND before the recipient needs to pass it on.

An object must be delivered from Person A to Person Z via a sequence of handovers. Several meetings are listed, but one crucial link is absent — a meeting between the person currently holding the object and the person who needs to pass it next. You pick which extra meeting would complete the chain.

Best approach: Draw the chain: A → B → ? → D → Z. Trace the object forward step by step, noting who holds it and when. Find the gap — the person who needs the object hasn't received it. The correct extra meeting must involve that person receiving the object from whoever currently holds it, at a time that still allows the chain to continue.

Question

Zara needs to deliver a birthday card to Will today. She will not see Will in person. These handovers are already scheduled:

  • Zara gives the card to Ben at 09:00.
  • Mia meets Cora at 11:00.
  • Cora meets Will at 13:00.
  • Ben meets Sam at 14:00.

Which one extra meeting, if added, could allow Will to receive the card today?

  1. ABen meets Zara at 07:30.
  2. BSam meets Cora at 10:00.
  3. CMia meets Ben at 08:30.
  4. DBen meets Mia at 10:00.

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Lesson 5 of 7Ordering DeductionIntermediate

Ordering problems do not always come as a list of finish positions. This one involves heights and stairs — and the key insight is that a staircase works like a measuring scale in reverse: taller people go on lower steps, shorter people go on higher steps, so everyone's head reaches the same level. Once you see that, the "impossible" option stands out immediately.

Height-compensation spatial ordering questions appear occasionally in OC TS and reward students who can translate a physical situation into a height ranking. The most common mistake is forgetting that the problem is underdetermined (we don't know Izzy vs Melissa) and wrongly ruling out options that are valid in one of the two possible scenarios.

The examiner is testing two skills at once: correctly ranking heights from relative clues (knowing that Caitlin is shortest, but that Izzy vs Melissa is unknown), and applying a staircase logic rule (shorter person → higher step) to determine which step assignments are forced and which are flexible.

Two or three clues establish a partial height order. A compensating physical setup (stairs, boxes, platforms) means each person's final head position depends on their natural height plus their step elevation. You must find the one step assignment that contradicts the forced height rank.

Best approach: First, extract everything the clues tell you — and be precise about what they do NOT tell you. Here, Caitlin is definitively the shortest, but the ordering of Izzy and Melissa is unknown. Second, apply the stair rule: shortest → top step, tallest → bottom step. The person who is definitively at one extreme (Caitlin = shortest) has a fixed position (top step). Any option that moves her out of that position cannot be true.

Question

Izzy is taller than Caitlin; Caitlin is shorter than Melissa.

But when they stand on a short flight of three stairs, one on each step, their heads are all level.

Which one of these sentences cannot be true?

  1. AIzzy is on the bottom step.
  2. BMelissa is on the middle step.
  3. CCaitlin is on the middle step.
  4. DMelissa is the tallest of the three girls.

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Lesson 6 of 7Ordering DeductionIntermediate

Idris has four treasure boxes (W, X, Y, Z) and three ordering clues. From them he can work out W > X > Z and W > Y — but Y's position relative to X and Z is still unknown (three possibilities). The question asks which fourth clue would make the ordering unique. The trick is to test each option and see whether it collapses all possibilities to exactly one.

'Find the completing clue' ordering questions appear occasionally in OC TS as a harder variant of standard linear ordering. They reward students who think systematically about gaps — what is not yet determined — rather than just what is already known. The most common mistake is picking option D (Y > Z), which narrows to two possibilities instead of one.

The examiner is testing whether students can identify what is already determined (W > X > Z, W > Y) and what remains undetermined (Y's position relative to X and Z), then find the single extra clue that resolves that ambiguity completely. Students who only check that an option is 'consistent' with the clues will incorrectly select B, C, or D.

Four items must be placed in a strict unique order. Three ordering clues are given. From them, a partial order is established but one item (Y) has multiple valid positions. Four options each propose a different fourth clue. Only one option, combined with the existing three clues, forces a unique complete ordering. The others either leave two orderings open or add a clue already implied by the existing set.

Best approach: Step 1: Chain the existing clues into a partial order. Step 2: Identify exactly what is unknown (which item has multiple valid positions). Step 3: For each option, add it to the partial order and check: does it leave exactly one possible full ordering? Reject any option that still allows two or more orderings. Reject any option already implied by the existing clues (it adds no information). The option that produces exactly one ordering is the answer.

Question

Idris is trying to solve a puzzle. He has four boxes labelled W, X, Y, and Z. He knows that there is a different amount of treasure in each box. He has four clues about the boxes (the fourth clue is not shown):

Clue 1: Box X contains more treasure than box Z. Clue 2: Box W contains more treasure than box Y. Clue 3: Box W contains more treasure than box X.

After reading all four clues, Idris is able to put the chests in order of how much treasure they contain.

Which one of the following could be the fourth clue?

  1. ABox Y contains more treasure than box X.
  2. BBox X contains more treasure than box Y.
  3. CBox W contains more treasure than box Z.
  4. DBox Y contains more treasure than box Z.

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Lesson 7 of 7Ordering DeductionDifficult

Five-person ordering puzzles are the flagship challenge of this unit — and they are far more systematic than they look. With five people and five clues, your brain wants to try every combination. But the right method cuts through all of that: anchor the one fixed position first, use chain clues to compress the remaining space, and treat adjacency pairs as locked units.

Five-person ordering questions with mixed relative, adjacency, and exclusion clues appear consistently in the difficult half of OC TS. They reward students who have a practised approach over those who rely on intuition. The same four clue types recur in almost every version: "last/first", "A before B", "immediately before", and "not first".

The examiner is testing whether you can simultaneously satisfy multiple ordering constraints — a fixed position anchor, a relative ordering chain, an adjacency constraint, and a position exclusion — and find the unique arrangement that satisfies all of them without contradiction.

Five people must be placed in positions 1–5. You get: one fixed-position clue (e.g. "Quinn is last"), one or two relative ordering chains (e.g. "Sasha before Ravi before Quinn"), one adjacency clue (e.g. "Tia immediately before Priya"), and one exclusion clue (e.g. "Tia is not first"). You are asked who occupies a specific position.

Best approach: Always start with the fixed-position clue — it gives you a definite anchor. Then expand the longest chain from that anchor to see how much space it consumes. Treat the adjacency pair as a single locked block and test which positions it can legally occupy given the exclusion. The one position set that satisfies every clue simultaneously is your answer.

Question

Five friends arrive at a cinema one at a time — no two arrive at the same moment. Their names are Priya, Quinn, Ravi, Sasha, and Tia.

  • Quinn arrives last.
  • Ravi arrives before Quinn but after Sasha.
  • Tia arrives immediately before Priya (no one else between them).
  • Tia does not arrive first.
  • Priya arrives before Ravi.

Who arrives second?

  1. APriya
  2. BRavi
  3. CSasha
  4. DTia

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