AMC 10 · 2017 · #14
Grade 4 countingPick an answer.
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.
Who is allowed beside Alice
Only two people may sit beside the tightest one.
Three people are banned from Alice's side, so only two are left to fill it.
4.OA.A.3Identify SubproblemsSplit by Alice's chair
That person's chair splits the count in two cases.
An end seat exposes Alice to one neighbor; an inner seat exposes her to two.
4.OA.A.3Draw A DiagramAlice at an end chair
An end chair has only one neighbour.
Pick Alice's one neighbor, arrange the rest, then drop the seatings where Derek and Eric collide.
4.NBT.B.5Make A Systematic ListAlice in a middle chair
A middle chair has two, which is tighter.
Sitting Alice between Derek and Eric satisfies every rule in one move.
4.NBT.B.5Make A Systematic ListAdd the two cases
Adding the cases gives 28, choice (C).
Separate, non-overlapping cases just add together.
Separate, non-overlapping cases just add together.
▸ Why?
Each seating falls into exactly one case, so nothing is counted twice and none is missed.
▸ Why?
Inside a case the remaining seats are filled independently, so those counts multiply first.
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