DILR guide, section 5

Arrangement and distribution

Both sets reward the same habit: place the most constrained item first, then let the remaining slots or totals decide the rest. Time yourself at 15 minutes per set before reading the solution.

Worked set 1, easy. Attempt it in 15 minutes first.

Circular arrangement

Six friends, Asha, Bala, Chirag, Divya, Esha and Farhan, sit around a round table, equally spaced and facing the centre.

  1. 1Asha sits directly opposite Divya.
  2. 2Bala sits immediately to the left of Asha.
  3. 3Esha does not sit next to Divya.
  4. 4Farhan does not sit next to Bala.

Questions

  1. Who sits directly opposite Bala?
  2. (TITA) Counting clockwise from Asha, how many people sit between Asha and Chirag?
  3. Who sits immediately to the right of Divya?
  4. (TITA) If clue 4 were removed, how many seating arrangements would satisfy the other clues?
Show solution

Number the seats 1–6 clockwise. Facing the centre, a person’s left neighbour is the next seat clockwise. Fix Asha at seat 1.

  • Clue 1: Divya at seat 4.
  • Clue 2: Bala at seat 2.
  • Seats 3, 5 and 6 remain for Chirag, Esha and Farhan. Clue 3: seats 3 and 5 touch Divya, so Esha takes seat 6.
  • Clue 4: seat 3 touches Bala, so Farhan takes seat 5 and Chirag seat 3.
Seat (clockwise)123456
PersonAshaBalaChiragDivyaFarhanEsha

Answers. (1) Farhan: seat 5 faces seat 2. (2) 1, namely Bala. (3) Chirag: Divya’s right is anticlockwise, seat 3. (4) 2: Esha stays at seat 6, and Chirag and Farhan can swap seats 3 and 5.

Takeaway. Fix one person, then use “opposite” and “immediately left/right” before any “not next to” clue. Question 4 is a typical CAT twist: drop a clue and count the cases.

15 minutes, timed

Practice set 1: linear arrangement

Seven students, P, Q, R, S, T, U and V, sit in a row of seats numbered 1 to 7 from left to right, all facing forward. “To the right of” means a higher seat number.

  1. 1P sits at one end of the row.
  2. 2Q sits in the middle seat.
  3. 3R sits immediately to the right of Q.
  4. 4Exactly two people sit between S and R.
  5. 5T sits immediately to the left of P.
  6. 6U does not sit next to Q.

Questions

  1. Who sits at the left end?
  2. (TITA) How many people sit between S and T?
  3. Who sits immediately to the left of Q?
  4. (TITA) If clue 6 were removed, how many arrangements would satisfy the other clues?
Show answers

(1) U. (2) 3. (3) V. (4) 2. The seating is 1 U, 2 S, 3 V, 4 Q, 5 R, 6 T, 7 P: Q takes seat 4 and R seat 5; S must be in seat 2; T needs a seat to P’s left, so P is 7 and T is 6; U avoids seat 3 (next to Q), so U is 1 and V is 3. Without clue 6, U and V can swap.

Worked set 2, medium. Attempt it in 15 minutes first.

Numeric distribution with cases

Four friends, A, B, C and D, bought 20 pens in all: 7 red, 8 blue and 5 green. Each bought at least one pen of each colour.

  1. 1A bought the most pens in total and D the fewest.
  2. 2B bought 3 red pens and equal numbers of blue and green pens.
  3. 3C bought exactly 5 pens, with more blue than red.
  4. 4D bought exactly 3 pens.

Questions

  1. (TITA) How many pens did B buy?
  2. Which of these must be true? (a) A bought 4 blue pens (b) C bought 2 green pens (c) A bought 2 red pens (d) C bought 3 blue pens
  3. (TITA) If A bought more blue pens than C, how many green pens did C buy?
  4. (TITA) How many different distributions satisfy all four clues?
Show solution

Write each person as (red, blue, green).

  • D: 3 pens with at least one of each, so D = (1, 1, 1).
  • C: 5 pens, at least one of each, blue > red. Only (1, 2, 2) and (1, 3, 1) work.
  • Red: A + 3 + 1 + 1 = 7, so A has 2 red pens in every case.
  • B = (3, x, x). A’s blue = 8 − x − C’s blue − 1 and A’s green = 5 − x − C’s green − 1, each at least 1.
CxBATotals A / B / C / DValid?
(1, 2, 2)1(3, 1, 1)(2, 4, 1)7/5/5/3Yes
(1, 2, 2)2(3, 2, 2)(2, 3, 0)—No: A has no green
(1, 3, 1)1(3, 1, 1)(2, 3, 2)7/5/5/3Yes
(1, 3, 1)2(3, 2, 2)(2, 2, 1)5/7/5/3No: A is not the most

Answers. (1) 5, in both valid cases. (2) (c). (3) 2: only the first case has A’s blue (4) above C’s (2). (4) 2.

Takeaway. Start with the most constrained person (D, then C), let the totals fix what they can, and list the surviving cases in one table. A question that adds a condition simply picks one of your cases.

15 minutes, timed

Practice set 2: numeric distribution

Three friends, X, Y and Z, share 18 chocolates: 9 dark and 9 milk. Each gets at least one of each kind.

  1. 1X gets twice as many chocolates in total as Z.
  2. 2Y gets equal numbers of dark and milk chocolates.
  3. 3Z gets more dark than milk chocolates.
  4. 4X gets at most 3 dark chocolates.

Questions

  1. (TITA) How many chocolates does Y get?
  2. (TITA) How many milk chocolates does X get?
  3. Which of these is true? (a) Z gets 2 milk chocolates (b) X gets 4 dark chocolates (c) Y gets 3 dark chocolates (d) Z gets 5 chocolates in all
  4. (TITA) What is the largest number of chocolates any one friend gets?
Show answers

(1) 6. (2) 5. (3) (c). (4) 8. With Z’s total z, X has 2z and Y has 18 − 3z. Y’s total is even, so z is even; Z needs at least 3 (two dark, one milk) and Y at least 2, so z = 4. Then Y has 6 (3 dark, 3 milk) and Z has 4, which must be 3 dark and 1 milk. That leaves X with 3 dark and 5 milk, which satisfies clue 4.