Arrangement sets in CAT LR: linear and circular seating, solved step by step

People in seats, books on shelves, cars in a parking row: an arrangement set asks you to place a group of entities in positions and match each one to an attribute. It is the most familiar logic type in CAT DILR, and the one where a clear diagram does most of the work.

The difficulty is rarely the logic. It is where you start, and how you handle a clue that allows two placements. This guide gives you a method and two original sets, a warm-up line and a CAT-style circle that needs a case split, each checked by computer to have exactly one solution.

How arrangement sets are asked

CAT sets usually add one or two attributes to the positions, and mix direct questions with conditional ones ("if clue 6 were removed…").

The method

1. Draw the positions before reading the clues closely: a row of numbered boxes, or a circle of numbered seats.
2. Sort the clues. Mark each as fixed ("P is in seat 5"), relative ("Q is immediately right of R"), or negative ("T is not next to Q").
3. Place fixed clues first. They give you anchors.
4. Attach relative clues to the anchors. "Immediately right of the dog owner" is useful only once you know where the dog owner is.
5. Split into cases when a relative clue has two possible placements. Draw each case as its own row and push both forward.
6. Use negative clues to kill cases. They rarely place anyone, but they often decide between two cases.
7. Check every clue against the final arrangement before answering.

Set 1: a linear warm-up

Five friends, P, Q, R, S and T, sit in seats numbered 1 to 5 from left to right. Each owns a different pet: cat, dog, fish, parrot or rabbit.

1. The dog owner sits in seat 3.
2. P sits at one end and owns the fish.
3. Q sits immediately to the right of the dog owner.
4. The rabbit owner sits in seat 2.
5. T owns the cat and does not sit next to Q.
6. R sits somewhere to the left of S.

Solution, clue by clue.

• Clues 1 and 4 fix two pets to seats: dog in seat 3, rabbit in seat 2.
• Clue 3 puts Q in seat 4.
• T owns the cat (clue 5), so T is not in seat 2 (rabbit) or seat 3 (dog). T is not next to Q, so T is not in seat 5. That leaves seat 1 for T.
• P is at an end (clue 2) and seat 1 is taken, so P is in seat 5 with the fish.
• R and S take seats 2 and 3. R is to the left of S (clue 6), so R is in seat 2 with the rabbit and S in seat 3 with the dog.
• The only pet left is the parrot, so Q owns the parrot.

Q1. Who owns the parrot? Q.

Q2. Who sits between R and Q? S.

Q3. Which pet does the person in seat 5 own? Fish.

Notice the order: fixed clues, then the relative clue that hangs on the dog, then a negative clue that forced T's seat.

Set 2: a circular set with a case split

Six friends, A, B, C, D, E and F, sit around a circular table in six evenly spaced seats numbered 1 to 6 clockwise, so seats 1 and 4, 2 and 5, and 3 and 6 face each other. Each friend is from a different city: Delhi, Jaipur, Kochi, Mumbai, Patna or Pune.

1. A sits in seat 1.
2. The friend from Kochi sits directly opposite A.
3. B sits immediately clockwise of the friend from Kochi.
4. C and F both sit next to A.
5. F is from Jaipur.
6. The friend from Delhi sits directly opposite C.
7. The friend from Pune sits next to B.
8. D is not from Kochi, and A is not from Patna.

Step 1: the anchors. A is in seat 1, so the Kochi friend is in seat 4 and B is in seat 5. C and F take seats 2 and 6, the two seats next to A. D and E take seats 3 and 4.

Step 2: split on C. Clue 4 allows two placements, so draw both.

Step 3: kill a case with clue 7. The Pune friend sits next to B, so in seat 4 or seat 6. Seat 4 is Kochi, so Pune is in seat 6.

• In Case I, seat 6 is F, but F is from Jaipur (clue 5). Case I fails.
• In Case II, seat 6 is C, so C is from Pune and F in seat 2 is from Jaipur. Case II survives.

Step 4: finish. C is in seat 6, so by clue 6 the Delhi friend is in seat 3. Seat 4 is Kochi. D is not from Kochi (clue 8), so D is in seat 3 (Delhi) and E in seat 4 (Kochi). A and B share Mumbai and Patna. A is not from Patna, so A is from Mumbai and B from Patna.

Q1. Who is from Delhi? D.

Q2. Who sits directly opposite F? F is in seat 2, and seat 5 faces it: B, from Patna.

Q3. Going clockwise from B, how many people sit between B and F? Seat 5 → seat 6 (C) → seat 1 (A) → seat 2 (F). Two: C and A.

Q4. If clue 8 were removed, how many complete arrangements (seats and cities) would be possible? D and E could swap seats 3 and 4, and A and B could swap Mumbai and Patna: 2 × 2 = 4.

Common mistakes

• Mixing up "immediately" and "somewhere". "R sits to the left of S" allows any gap; "immediately to the left" allows none.
• Assuming a direction in a circle. State the convention on your rough sheet. If people face the centre, their right hand points anticlockwise when you look down on the table. This set uses clockwise seat numbers so that no convention is needed.
• Forgetting that "next to" has two sides. In a circle, every seat has two neighbours; at the end of a row, only one.
• Dropping a case too early. Case I in set 2 looked fine until clue 7. Keep both rows until a clue actually breaks one.

Practice set

1. In set 1, suppose clue 6 (R to the left of S) is removed. How many arrangements are possible, and which pet is still definitely in seat 3?
2. In set 2, suppose clue 7 (Pune next to B) is removed but every other clue stays. Which friend's seat and city are both still fixed?
3. In set 2, which single clue rules out Case I?

Answers:

1. Two arrangements; the dog is still in seat 3. R and S can swap seats 2 and 3, but the pets are tied to seats, so only the dog's owner changes. A seat can be fixed without its person being fixed.
2. E only (seat 4, Kochi). Without clue 7, both cases survive. In Case I, B (seat 5, opposite C) is from Delhi; in Case II, D is. Either way D is in seat 3 and, since D is not from Kochi, E is the Kochi friend in seat 4. D, B and A have fixed seats but open cities, and C and F can swap seats. A computer check of all eight surviving arrangements confirms it.
3. Clue 7, together with clue 5. Clue 7 puts Pune in seat 6, and in Case I that seat belongs to F, who is from Jaipur.

What to do next

• Solve five linear and five circular sets this week, drawing the empty positions before you read the first clue.
• On every set, label each clue fixed, relative or negative before placing anything.
• Whenever a clue allows two placements, draw both cases as separate rows, as in set 2.
• Move on to scheduling sets, which use the same method with days and slots instead of seats, and to choosing which sets to attempt.