AMC 10 · 2018 · #18

Grade 7 arithmetic

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

permutations-basicsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: permutations-basic

📏 Medium solution 💡 3 insights

Problem

Six children — three brother-sister pairs from three different families — fill two rows of three seats each (a 2× 3 grid of seats). Two rules apply: siblings may not sit side by side in the same row, and no child may sit directly in front of or behind a sibling (same column). Count how many seating arrangements obey both rules.

Pick an answer.

(A)

$text{ 60}$

(B)

$text{ 72}$

(C)

$text{ 92}$

(D)

$text{ 96}$

(E)

$text{ 120}$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Make a Systematic List): there are only a handful of valid seating shapes, so the safe way to get an exact count is to list them in an organized order that can't skip a case or count one twice. Tool #1 (Draw a Diagram): drawing the 2× 3 grid turns the two rules into simple statements about which seats a family may share. Tool #7 (Identify Subproblems): split the job into 'which family sits in each seat' first, then 'which sibling of the family takes which seat' — two easy counts multiplied together. Tool #3 (Eliminate Possibilities): for each partly-filled grid, most ways to continue break a rule, so crossing those out quickly leaves the valid ones.

1STEP 1

Draw the grid and translate the rules

Sketch a 2× 3 grid; the two rules mean a family's two seats must be in different columns and never same-row neighbors.

F₁ & F₂ & F₃ ; B₁ & B₂ & B₃ ;

2STEP 2

Split into two easier counts

Count in two stages: first the family pattern (who sits where), then 2³ ways to swap siblings inside families; multiply the two counts.

total = (# valid family patterns) × 2³

3STEP 3

List the valid patterns with family A at front-left

Anchor family A in seat F₁; its partner must be B₂ or B₃, and filling B and C the only legal ways gives exactly 4 patterns.

front A B C & & front A C B ; back B C A & & back C B A front A B C & & front A C B ; back C A B & & back B A C

4STEP 4

Account for which family sits front-left

Family B or C could instead sit in F₁; the three non-overlapping groups give 4 × 3 = 12 valid family patterns.

4 × 3 = 12 valid family patterns

5STEP 5

Bring back the two siblings per family

Each of the 12 patterns splits into 2³ = 8 real seatings, so 12 × 8 = 96 arrangements obey both rules — choice (D).

12 × 2³ = 12 × 8 = 96 → (D)

Answer

text{ 96}

A quick divisibility filter confirms the size. Swapping the two siblings inside each of the three families multiplies any valid count by 2³ = 8, and relabeling the three families among their seat-pairs multiplies it by 3! = 6; since no valid layout is left unchanged by either operation, the true answer must be a multiple of 8 × 6 = 48. Among the choices 60, 72, 92, 96, 120, the only multiple of 48 is 96. This both pins the answer to (D) and matches our direct count 12 × 8 = 96.

💡Key takeaway

Count the family layouts first (there are 12), then multiply by 2³=8 for swapping the two siblings inside each family to get 96.

Draw the grid and translate the rules
Split into two easier counts
List the valid patterns with family A at front-left
Account for which family sits front-left
Bring back the two siblings per family

Draw the grid and translate the rules

Split into two easier counts

List the valid patterns with family A at front-left

Account for which family sits front-left

Bring back the two siblings per family

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