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UPSC CSE · CSAT · Reasoning and judgment · TOPIC LESSON

Ordering grouping and arrangements

Translate arrangement rules exactly, enumerate valid cases and separate fixed facts from remaining choices.

UPSC Civil Services40 min study guide8 indexed sections

BY THE END, YOU CAN

  • Solve linear, circular, grouping and scheduling constraints.
  • Explain rank and inward-facing conventions.
  • Use complete case lists to justify must, could and cannot.

BEFORE YOU START

  • Distinguish a strict implication from its converse.
  • Read left-to-right positions as numbered slots.

Original educational lesson. Work through the examples before revealing the practice answers. This original common-core lesson supports its declared paper scope. Optional subjects, qualifying language papers and the interview are outside this resource.

LEARN THROUGH ORIGINAL PRACTICE

Original UPSC-style worked applications and practice

All teaching examples, cases, prompts and model answers are original. Official exam questions and answer keys are not reproduced. This common-core lesson does not imply complete historical paper coverage or an official score.

01 / UNDERSTAND & APPLY

Translate before arranging

Start with the objects, the available positions and their distinctness. Four distinct people in four slots means each person occupies one slot and no slot is shared. “A before B” permits intervening people; “A immediately before B” makes the ordered block AB. “A beside B” permits either AB or BA.

Group membership is not order. If A and B must share a team, they need not stand next to each other. Capacity also needs precision: at most three permits fewer, whereas exactly three does not. Write these distinctions before trying a promising arrangement.

A useful case table separates fixed information from a trial choice. Check each completed row against every rule, rather than continuing from an early guess as though it were established.

ORIGINAL PRACTICE · 1

Four books W, X, Y and Z occupy four distinct left-to-right positions. W is immediately before X and Y is immediately before Z. List the valid orders.

Reveal answer & reasoning

WXYZ and YZWX.

Treat WX and YZ as two ordered blocks. The blocks can appear in either order, but reversing a block would violate its immediate-before rule. No other slot remains between members of either block.

02 / UNDERSTAND & APPLY

Use blocks and an exhaustive slot list

A strong adjacency constraint can reduce a problem’s size. Treat an ordered pair as a block, but remember the block still occupies two physical slots. Enumerate possible block starts, fill the remaining positions under the other rules, and check whether any start is impossible.

Do not use a single valid row as proof of uniqueness. A complete list requires either systematic enumeration or an argument excluding every other case. If two rules force incompatible positions, report inconsistency rather than quietly dropping a rule.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Complete valid row arrangementsRows ABCD, CABD and CDAB show every allowed placement of the ordered AB block with C left of D.Slot 1Slot 2Slot 3Slot 4Case 1ABCDCase 2CABDCase 3CDABAB remains adjacent and ordered; C is left of D.
Original finite-case table. Shaded AB cells mark the ordered block, not extra constraints.

ORIGINAL WORKED EXAMPLE

Three possible rows

A, B, C and D occupy four distinct left-to-right slots. A is immediately before B, and C is somewhere left of D. List all valid arrangements.

  1. The ordered block AB can begin in slot 1,2 or 3. These exhaust its possible starts.
  2. If AB occupies 1–2, the remaining slots 3–4 must contain C then D: ABCD.
  3. If AB occupies 2–3, remaining slots 1 and 4 contain C then D: CABD.
  4. If AB occupies 3–4, remaining slots 1–2 contain C then D: CDAB. Each row satisfies both rules.

Answer ABCD, CABD and CDAB are the complete valid list. An order such as ACDB fails because A and B are not adjacent.

ORIGINAL PRACTICE · 1

In the row problem with valid arrangements ABCD, CABD and CDAB, is “B immediately precedes C” necessary, possible or impossible?

Reveal answer & reasoning

Possible, but not necessary.

It occurs in ABCD. It fails in CABD and CDAB, both of which satisfy the original rules. One witness proves possibility; the other valid rows disprove necessity.

03 / UNDERSTAND & APPLY

Ranks count from a stated end

For one person with distinct ranks, position p from the front and q from the back imply total p+q−1. The person is counted in both p and q, so subtract one duplicate. This formula refers to the same person in the same line, with no ties or unfilled positions.

For people at positions p and q counted from the same end, the number strictly between them is |p−q|−1 when they are distinct. The position difference counts the step into the later person’s place; subtract that endpoint. Do not combine opposite-end ranks without first finding a common reference.

ORIGINAL WORKED EXAMPLE

Noor’s place in the row

A line has no ties. Noor is second from the left and fourth from the right. Mina is fifth from the left. How many people are in the line, and how many stand between Noor and Mina?

  1. Use Noor’s two ranks for the same person:2+4−1=5 people.
  2. Number from the left: Noor is 2 and Mina is 5.
  3. Their position difference is 3; the strictly intervening slots are 3 and 4, so there are 2 people.

Answer Five people are in the line, and two stand between Noor and Mina.

ORIGINAL PRACTICE · 1

In a line of ten people with distinct positions, someone is fourth from the front. What is their rank from the back?

Reveal answer & reasoning

Seventh.

Front rank plus back rank minus one equals 10, so back rank=10−4+1=7. Counting the person in both directions explains the added one.

ORIGINAL PRACTICE · 2

Two people are third and eighth from the same end of a line. How many people are strictly between them?

Reveal answer & reasoning

Four.

The intervening positions are 4,5,6 and 7. The calculation is 8−3−1=4, not the raw position difference 5.

04 / UNDERSTAND & APPLY

Fix a reference and a facing convention

If circular arrangements differing only by rotation count as the same, fix one person at a display position to remove duplicate rotations. This is a bookkeeping choice, not a new compass constraint. Reflection may still produce a different arrangement because it reverses left and right.

For people facing inward, viewed from above, a person’s immediate left is the clockwise neighbouring position, and immediate right is anticlockwise. Check at the top seat: the person faces down toward the centre, so their left points toward the page’s right. If people face outward, the convention reverses.

State the facing rule and keep it consistent. “Left on the page” is not the same as “to a seated person’s left”. When an actual north position is part of the problem, absolute orientation cannot simply be discarded.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

An inward-facing circleA is at the top, B right, C bottom and D left, each facing the centre. From A, left is B and right is D. From B, left is C.ABCDTop view; everyone faces the centreCentreClockwise A → B → C → D is inward-facing left.
A is fixed at the top only to remove equivalent rotations; this is a display reference, not an added compass constraint.

ORIGINAL WORKED EXAMPLE

Four inward-facing seats

A, B, C and D sit at four equally spaced positions around a circle, all facing inward. Arrangements equivalent under rotation count as the same. B sits immediately to A’s left, and C sits opposite A. Find D’s relation to A.

  1. Fix A at the top display position. Since A faces the centre, A’s left is the right-hand position on the page. Place B there.
  2. Place C opposite A at the bottom. The only remaining position is the left-hand position on the page for D.
  3. Relative to inward-facing A, that position is A’s right. Check the facing direction rather than the viewer’s left/right labels.

Answer D sits immediately to A’s right. The displayed order clockwise is A, B, C, D.

ORIGINAL PRACTICE · 1

A, B, C and D occupy top, right, bottom and left positions respectively, all facing inward. Who is immediately to B’s left?

Reveal answer & reasoning

C.

B faces left toward the centre. B’s left is downward on the page, which is C’s position. Equivalently, inward-facing left follows the clockwise sequence.

05 / UNDERSTAND & APPLY

Membership under exact capacities

For labelled teams, exchanging team names usually creates a different assignment because the labels can carry different constraints. Do not divide a count by two unless the groups really are interchangeable. A fixed member in Red makes that label particularly important.

Use together and apart constraints as membership rules. If A and B are together, decide which team contains the pair, then fill capacities. A rule C not with A also separates C from B in that case. Distinguish this consequence from an unrelated ban on C meeting B in another task.

ORIGINAL WORKED EXAMPLE

Two teams of exactly three

Six distinct people A, B, C, D, E and F form labelled teams Red and Blue, each with exactly three people. A and B must be together, C must not be with A, and D must be in Red. List all possible Red teams.

  1. Case 1: A and B are in Red. D is also fixed Red, so Red=ABD and Blue=CEF. All sizes and separations hold.
  2. Case 2: A and B are in Blue. C cannot join them, so C is in Red with D.
  3. The third Red member is E or F. These yield Red=CDE, Blue=ABF, or Red=CDF, Blue=ABE.
  4. The pair AB must be in one of the two teams, so the cases are exhaustive.

Answer The possible Red teams are ABD, CDE and CDF, with their stated complementary Blue teams.

ORIGINAL PRACTICE · 1

Under an exactly-three-per-team requirement, someone proposes Red=ABDE and Blue=CF. Is it valid?

Reveal answer & reasoning

No. Red has four members and Blue has two.

Exactly three is an equality condition for each team. Red violates even a capacity-three limit, and Blue also violates the stated minimum implicit in exactly three. Under merely at-most-three wording, a two-member team alone would not be a violation.

06 / UNDERSTAND & APPLY

Order constraints are not full schedules

A before B imposes a precedence relation, not necessarily consecutive slots. Several precedence relations can form a partial order that still permits multiple full schedules. A fixed-day condition supplies different information from a relative-order condition.

Look for tasks that must occur after several others; this can force a late slot. Then enumerate the remaining tasks. Check that your final schedule respects both the fixed positions and every before/after relation.

ORIGINAL WORKED EXAMPLE

Two schedules survive

Tasks P, Q, R and S occupy days 1–4, one distinct task per day. P is before R, S is before R, and Q is fixed on day 2. List the schedules.

  1. R must have both P and S before it, so it cannot occupy day 1 or 2.
  2. If R were on day 3, the only earlier days 1 and 2 would need P and S, but day 2 is occupied by Q. Thus day 3 is impossible.
  3. R is on day 4. P and S fill days 1 and 3 in either order, while Q remains day 2.

Answer P,Q,S,R and S,Q,P,R. Q and R have fixed days; P and S remain interchangeable under the given rules.

ORIGINAL PRACTICE · 1

Three tasks P, Q and R occupy three days. P is before Q and R is before Q. Which task is necessarily last, and what are the schedules?

Reveal answer & reasoning

Q is last; the schedules are P,R,Q and R,P,Q.

Both other tasks must precede Q, which forces the third slot. Their order is not constrained, leaving two valid possibilities.

07 / UNDERSTAND & APPLY

Apply an if-rule only where it applies

An implication constrains cases where its antecedent is true. If A first implies B second, arrangements with A elsewhere are not restricted by that rule. Requiring B second in every case would strengthen the premise incorrectly.

Enumerate the base arrangements, then test the conditional. Keeping cases separate helps prevent duplicate counting and accidental carry-over from one assumed position to another. If a case begins with a temporary assumption, label it so later reasoning does not treat it as a global fact.

ORIGINAL PRACTICE · 1

A, B and C occupy days 1–3. The only extra rule is: if A is first, B is second. List all permitted orders.

Reveal answer & reasoning

ABC, BAC, BCA, CAB and CBA.

The six unrestricted permutations are ABC, ACB, BAC, BCA, CAB, CBA. ACB alone has A first without B second, so it is excluded. In the other four non-A-first cases, the antecedent is false and the rule adds no restriction.

08 / UNDERSTAND & APPLY

Report exactly what the cases establish

An invariant is a feature shared by every valid arrangement. In the schedule example, R on day 4 is invariant even though two full schedules remain. In the grouping example, D in Red is a given invariant, while C in Red is not: the ABD case places C in Blue.

For could-be questions, exhibit one complete valid case. For must-be questions, use the whole valid set or a proof that excludes every alternative. For cannot-be questions, show why the proposed condition conflicts with a rule in every possible case.

Before finalizing, count objects and slots, verify capacities, check direction/facing and test each original constraint. If no case survives, reconsider your translation; if the premises really conflict, report inconsistency rather than invent a repaired version.

KEEP THE BIG IDEAS

Quick revision

  • Immediate order, general order, adjacency and membership are different constraints.
  • Exact capacities are stronger than upper limits.
  • Use one consistent rank or facing convention.
  • An implication does not constrain cases where its antecedent is false.
  • A full case list can establish invariants without yielding a unique arrangement.
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