AMC 10 · 2020 · #15

Grade 4 counting

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

systematic-enumerationcombinations-basicsymmetry-argument caseworkidentify-subproblemssystematic-enumeration ↑ Prerequisites: systematic-enumerationcombinations-basic

📏 Medium solution 💡 2 insights

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Problem

Ten people stand equally spaced around a circle. Each knows exactly three others: their two neighbours plus the one directly across the circle. Count the ways to split the ten into five pairs so that every pair is made of people who know each other.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): draw the 10-cycle with the 5 diameters across it — the picture instantly shows the two pair types (neighbor edge or diameter). Tool #7 (Subproblems): split into cases by how many diameters k the pairing uses (k = 0, 1, 2, 3, 4, 5); inside each case the remaining people must pair as neighbors. Tool #2 (Systematic List): inside each case, list the configurations in order so nothing is missed and nothing is double-counted.

1STEP 1

Split into cases

Case on the number of across-pairs.

cases by k = 0, 1, 2, 3, 4, 5

2STEP 2

All across-pairs

There is exactly one such split.

#{k = 5} = 1

3STEP 3

Four across-pairs

The last two would not be neighbours.

#{k = 4} = 0

4STEP 4

Three across-pairs

The other four pair as neighbours.

#{k = 3} = 5

5STEP 5

Two across-pairs

This case is impossible too.

#{k = 2} = 0

6STEP 6

One across-pair

Both arcs fill with neighbour pairs.

#{k = 1} = 5 · 1 · 1 = 5

7STEP 7

No across-pairs

Two ways to fill the circle by alternating.

#{k = 0} = 2

8STEP 8

Add the cases

Adding gives 13.

1 + 0 + 5 + 0 + 5 + 2 = 13 → (C)

Answer

The answer 13 sits in the middle of the offered range 11 to 15, exactly where casework with two large contributing buckets (the k = 1 and k = 3 cases give 5 each) plus the small k = 0 and k = 5 buckets (2 + 1 = 3) lands. The two impossible cases (k = 2 and k = 4) make sense from the picture: the leftover arcs have odd length or strand opposite endpoints, so no neighbor matching exists.

💡Key takeaway

This AMC 12 problem only needs Grade 4 case-by-case counting you already know — draw the 10 people around the circle and split by how many cross-diameter pairs the matching uses (k = 0, 1, 2, 3, 4, 5). The cases give 2 + 5 + 0 + 5 + 0 + 1 = 13 valid pairings. The answer is (C).