Competition · AMC preparation · step 4 of 4

AMC 10 · 2020B · #17

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 3 others: their 2 neighbors plus the one directly across the circle. Count the ways to split the 10 into 5 pairs so that every pair is made of people who know each other.

Pick an answer.

(A)
11
(B)
12
(C)
13
(D)
14
(E)
15

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Label the ten people

Label people 0–9; every pair is a neighbor edge or a diameter, so split cases by the number of diameters k the matching uses.

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

Count the five-diameter case

Case k = 5: all five diameters used, everyone paired across the circle — exactly 1 way.

#{k = 5} = 1
3STEP 3

Count the four-diameter case

Case k = 4: the two leftover people are opposite ends of the unused diameter, not neighbors, so they cannot pair — impossible, 0 ways.

#{k = 4} = 0
4STEP 4

Count the three-diameter case

Case k = 3: works only when the two skipped diameters are adjacent on the diameter-cycle of 5, giving 5 ways.

#{k = 3} = 5
5STEP 5

Count the two-diameter case

Case k = 2: leftover people form two arcs of 3, and 3 in a row has no perfect neighbor matching — impossible, 0 ways.

#{k = 2} = 0
6STEP 6

Count the one-diameter case

Case k = 1: pick 1 of 5 diameters, and each leftover arc of 4 pairs uniquely as neighbors — 5 ways.

#{k = 1} = 5 · 1 · 1 = 5
7STEP 7

Count the no-diameter case

Case k = 0: all pairs are neighbor edges, and a 10-cycle has exactly 2 ways to match perfectly.

#{k = 0} = 2
8STEP 8

Add up the cases

Add the surviving cases: 1 + 5 + 5 + 2 = 13, which is choice (C).

1 + 0 + 5 + 0 + 5 + 2 = 13 → (C)
Answer
13
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 10 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).

  • Label the ten people
  • Count the five-diameter case
  • Count the four-diameter case
  • Count the three-diameter case
  • Count the two-diameter case
  • Count the one-diameter case
  • Count the no-diameter case
  • Add up the cases

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