AMC 10 · 2020 · #15
Easy mode Grade 4Picture 10 people standing in a circle, spaced evenly apart.
Each person knows exactly 3 of the other 9 people:
- the person standing on their left,
- the person standing on their right,
- and the person standing directly across the circle from them.
We want to split the 10 people into 5 pairs. In every pair, the two people must already know each other.
How many different ways can we do this?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: $10$ people at equally spaced positions around a circle (label them $0, 1, \dots, 9$); Person $i$ knows person $i - 1$ and person $i + 1$ (neighbors, indices mod $10$) and person $i + 5$ (directly across, a 'diameter'); There are exactly $5$ diameters: $(0,5), (1,6), (2,7), (3,8), (4,9)$; Choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Unknowns: Number of valid $5$-pair groupings of the $10$ people
Understand
Restated: 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.
Givens: $10$ people at equally spaced positions around a circle (label them $0, 1, \dots, 9$); Person $i$ knows person $i - 1$ and person $i + 1$ (neighbors, indices mod $10$) and person $i + 5$ (directly across, a 'diameter'); There are exactly $5$ diameters: $(0,5), (1,6), (2,7), (3,8), (4,9)$; Choices: (A) $11$, (B) $12$, (C) $13$, (D) $14$, (E) $15$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
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.
Execute — Answer: C
4.OA.C.5 Step 1 - Label the people $0, 1, \dots, 9$ around the circle.
- Each pair must be a neighbor edge or a diameter.
- Split the count by $k$, the number of diameter pairs used.
- The non-diameter people must pair up among themselves using only neighbor edges.
💡 Sort the configurations by how many cross-diameters they use.
4.OA.C.5 Step 2 - Case $k = 5$ (all five diameters).
- Every person is paired across the circle.
- Exactly $1$ configuration.
💡 All diameters used — only one way.
4.G.A.1 Step 3 - Case $k = 4$.
- Use four diameters; the two leftover people are the endpoints of the unused diameter — they are directly across each other, not neighbors.
- They cannot pair (the diameter is excluded by assumption and they aren't adjacent).
- Impossible.
- $0$ ways.
💡 Dropping one diameter strands two opposite people with no legal edge between them.
4.OA.C.5 Step 4 - Case $k = 3$.
- Use three diameters; the four leftover people fill two opposite arcs and must pair as neighbors.
- This works iff the two omitted diameters are 'adjacent' (consecutive among the five diameters), so each side has two neighboring people.
- The number of adjacent pairs of omitted diameters around the diameter-cycle of length $5$ is $5$.
💡 The two skipped diameters must be next to each other so the leftover people land in neighbor pairs.
4.G.A.1 Step 5 - Case $k = 2$.
- Two diameters used, six leftover people in two opposite arcs of $3$ each.
- An arc of three consecutive people on the cycle has no perfect neighbor matching (any matching leaves one person out).
- Impossible.
- $0$ ways.
💡 Three people in a row can't pair off using only neighbor edges.
4.OA.C.5 Step 6 - Case $k = 1$.
- Choose $1$ of $5$ diameters ($5$ choices).
- The eight leftover people sit on two opposite arcs of $4$ consecutive people each.
- Each arc of $4$ has exactly $1$ neighbor-pair perfect matching.
- Total $5 \cdot 1 = 5$.
💡 Pick the single diameter, then the two arcs of $4$ each pair up uniquely as neighbors.
4.OA.C.5 Step 7 - Case $k = 0$ (no diameters).
- All five pairs are neighbor edges on the $10$-cycle.
- A $10$-cycle has exactly $2$ perfect matchings: $(0,1)(2,3)(4,5)(6,7)(8,9)$ and $(1,2)(3,4)(5,6)(7,8)(9,0)$.
💡 Two ways to pair $10$ chairs around a round table — even-odd or odd-even.
1.OA.A.2 Step 8 - Add the contributing cases: $1 + 5 + 5 + 2 = 13$.
- That matches choice (C).
💡 Disjoint cases — just add.
4.OA.C.5 Label the people $0, 1, \dots, 9$ around the circle. Each pair must be a neighbo 4.OA.C.5 Case $k = 5$ (all five diameters). Every person is paired across the circle. Exa 4.G.A.1 Case $k = 4$. Use four diameters; the two leftover people are the endpoints of t 4.OA.C.5 Case $k = 3$. Use three diameters; the four leftover people fill two opposite ar 4.G.A.1 Case $k = 2$. Two diameters used, six leftover people in two opposite arcs of $3 4.OA.C.5 Case $k = 1$. Choose $1$ of $5$ diameters ($5$ choices). The eight leftover peop 4.OA.C.5 Case $k = 0$ (no diameters). All five pairs are neighbor edges on the $10$-cycle 1.OA.A.2 Add the contributing cases: $1 + 5 + 5 + 2 = 13$. That matches choice (C). Review
Reasonableness: 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.
Alternative: Tool #2 (Systematic List) directly: list every perfect matching of the graph with vertices $0$-$9$ and edges $\{i, i+1\}$ and $\{i, i+5\}$. Order the matchings by which edge contains vertex $0$ (three choices: $\{0,1\}, \{0,9\}, \{0,5\}$) and recursively pair the remaining vertices. Counting yields the same $13$.
CCSS standards used (min grade 4)
1.OA.A.2Solve word problems involving three whole numbers whose sum is within 20 (Adding the contributions $1 + 5 + 5 + 2 = 13$ across cases.)4.OA.C.5Generate a number or shape pattern following a given rule (Counting configurations in each case by following the rule 'every pair is a neighbor edge or a diameter'.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Drawing the $10$ points around the circle with the $5$ diameters to see which leftover arcs can be perfectly matched by neighbors.)
⭐ 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 $\textbf{(C)}$.
⭐ 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 $\textbf{(C)}$.
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