AMC 10 · 2017 · #19

Grade 4 counting
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
permutations-basiccasework caseworkcomplementary-countingprinciple-of-inclusion-exclusion ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 3 insights
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Problem
Five people — Alice, Bob, Carla, Derek, and Eric — sit in a row of 5 chairs. Alice will not sit directly next to Bob or Carla, and Derek will not sit directly next to Eric. Count the number of seatings that obey both rules.

Pick an answer.

(A)
12
(B)
16
(C)
28
(D)
32
(E)
40

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question asks "how many ways," so Tool #2 (Make a Systematic List) fits: build the seatings in an organized order instead of guessing. The whole problem hinges on Alice, because three of the five people are banned from her side. Tool #7 (Identify Subproblems) splits the count by where Alice sits — at an end (one neighbor) or in the middle (two neighbors) — since those two situations behave very differently. Tool #1 (Draw a Diagram) — five chairs in a row — keeps "neighbor" concrete so each placement is easy to check.

1STEP 1

Who is allowed beside Alice

Bob and Carla are banned from Alice's side, so only Derek and Eric may sit beside her; organize the count by how many neighbors Alice has.

neighbor of Alice ∈ {Derek, Eric}
2STEP 2

Split by Alice's chair

An end chair (1 or 5) has one neighbor; each inner chair (2, 3, 4) has two — so count end and inner seatings separately, then add.

ends: chairs 1,5 middle: chairs 2,3,4
3STEP 3

Alice at an end chair

With Alice at an end, pick her neighbor (Derek or Eric), arrange the rest, and drop the Derek–Eric collisions: 8 per end, so 16 total.

2 × (6-2) = 8 per end, 8 × 2 = 16
4STEP 4

Alice in a middle chair

With Alice inner, Derek and Eric must take both sides (auto-satisfying their rule); Bob and Carla fill the rest: 4 each, 12 total.

2 × 2 = 4 per inner chair, 4 × 3 = 12
5STEP 5

Add the two cases

The end and inner cases never overlap, so add: 16 + 12 = 28, which is choice (C).

16 + 12 = 28 → (C)
Answer
28
Without any rules, five people seat in 5!=120 ways, so a valid count of 28 is a sensible fraction of that — small but not tiny, matching how strong the restrictions are. The two cases cover every position for Alice (two ends plus three inner chairs) with no overlap, so nothing is missed or counted twice. The total 28 is exactly choice (C).
💡Key takeaway

When one person blocks most neighbors, count by where that person sits, and the rest falls into a few easy cases.

  • Who is allowed beside Alice
  • Split by Alice's chair
  • Alice at an end chair
  • Alice in a middle chair
  • Add the two cases

More like this

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