AMC 10 · 2017 · #19
Easy mode Grade 4Five friends — Alice, Bob, Carla, Derek, and Eric — sit in a row of 5 chairs. Alice will not sit right next to Bob, and she will not sit right next to Carla. Derek will not sit right next to Eric. In how many ways can the five of them sit in the row while following all these rules?
Pick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: There are $5$ people and $5$ chairs in a single row; Alice refuses to sit immediately next to Bob; Alice refuses to sit immediately next to Carla; Derek refuses to sit immediately next to Eric; Answer choices: (A) $12$, (B) $16$, (C) $28$, (D) $32$, (E) $40$
Unknowns: The number of valid orderings of the five people in the row
Understand
Restated: 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.
Givens: There are $5$ people and $5$ chairs in a single row; Alice refuses to sit immediately next to Bob; Alice refuses to sit immediately next to Carla; Derek refuses to sit immediately next to Eric; Answer choices: (A) $12$, (B) $16$, (C) $28$, (D) $32$, (E) $40$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #1 Draw a Diagram
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.
Execute — Answer: C
4.OA.A.3 Step 1 Who is allowed beside Alice
- Alice cannot sit next to Bob or Carla.
- The only people left who may sit directly beside her are Derek and Eric.
- So every chair touching Alice must hold Derek or Eric.
- This one fact controls the whole count, so organize the cases by how many neighbors Alice has.
💡 Three people are banned from Alice's side, so only two are left to fill it.
4.OA.A.3 Step 2 Split by Alice's chair
- An end chair (chair $1$ or chair $5$) has just one neighbor.
- Each of the three inner chairs ($2$, $3$, $4$) has two neighbors.
- Because the allowed-neighbor rule depends on how many neighbors Alice has, count the end-chair seatings and the inner-chair seatings separately, then add them.
💡 An end seat exposes Alice to one neighbor; an inner seat exposes her to two.
4.NBT.B.5 Step 3 Alice at an end chair
- Put Alice at chair $1$.
- Her single neighbor (chair $2$) must be Derek or Eric: $2$ choices.
- The remaining three people — Bob, Carla, and whichever of Derek/Eric is left — fill chairs $3,4,5$ in $3\times2\times1=6$ orders.
- But Derek and Eric must not touch: that happens only when the leftover Derek/Eric lands in chair $3$ (next to the one in chair $2$), which is $1\times2=2$ bad orders.
- So $6-2=4$ good orders.
- That gives $2\times4=8$ seatings for chair $1$, and chair $5$ is the mirror image, so the two end chairs give $8\times2=16$.
💡 Pick Alice's one neighbor, arrange the rest, then drop the seatings where Derek and Eric collide.
4.NBT.B.5 Step 4 Alice in a middle chair
- Now put Alice in an inner chair ($2$, $3$, or $4$).
- She has two neighbors, and both must come from {Derek, Eric}.
- So Derek and Eric sit on her two sides in $2$ orders.
- Since Alice sits between them, Derek and Eric are automatically not next to each other — that rule takes care of itself.
- Bob and Carla fill the last two chairs in $2$ orders.
- Each inner chair gives $2\times2=4$ seatings, and there are three inner chairs, so $4\times3=12$.
💡 Sitting Alice between Derek and Eric satisfies every rule in one move.
3.NBT.A.2 Step 5 Add the two cases
- The end-chair seatings and the inner-chair seatings never overlap, so add them: $16+12=28$.
- That is choice (C).
💡 Separate, non-overlapping cases just add together.
4.OA.A.3 Alice cannot sit next to Bob or Carla. The only people left who may sit directly 4.OA.A.3 An end chair (chair $1$ or chair $5$) has just one neighbor. Each of the three i 4.NBT.B.5 Put Alice at chair $1$. Her single neighbor (chair $2$) must be Derek or Eric: $ 4.NBT.B.5 Now put Alice in an inner chair ($2$, $3$, or $4$). She has two neighbors, and b 3.NBT.A.2 The end-chair seatings and the inner-chair seatings never overlap, so add them: Review
Reasonableness: 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).
Alternative: Use complementary counting with inclusion-exclusion. From $120$ total, subtract the seatings that break a rule: Alice-by-Bob, Alice-by-Carla, and Derek-by-Eric each give $2\cdot4!=48$. Add back the overlaps (Alice by both Bob and Carla; Alice-Bob with Derek-Eric; Alice-Carla with Derek-Eric), then subtract the triple overlap. The inclusion-exclusion total of forbidden seatings comes to $92$, and $120-92=28$, confirming (C).
CCSS standards used (min grade 4)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Turning the seating rules into a plan: deduce who may sit beside Alice, then split the count by where Alice sits.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Multiplying the number of choices in each case — neighbor choice times arrangements — to count seatings.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the end-chair count and the inner-chair count to reach $16+12=28$.)
⭐ When one person blocks most neighbors, count by where that person sits, and the rest falls into a few easy cases.
⭐ When one person blocks most neighbors, count by where that person sits, and the rest falls into a few easy cases.
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