AMC 10 · 2013 · #15

Grade 7 counting
Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type
fundamental-counting-principlecaseworkexponents identify-subproblemssystematic-enumeration ↑ Prerequisites: fundamental-counting-principle
📏 Medium solution 💡 3 insights
Problem
Two groups are sent to several places and no place may hold members of both. Count the ways.

Pick an answer.

(A)
96
(B)
108
(C)
156
(D)
204
(E)
372
How to solve
Strategy Identify Subproblems

Tool #16 (Change Focus) comes first, because the ban is stated as four separate store conditions and the whole problem opens up once those collapse into a single statement about the parents' stores. That reframing shows the children's options depend on one number only: how many distinct stores the parents fill. Tool #7 (Identify Subproblems) is primary because that number can only be 1 or 2, which gives a case split that is both exhaustive and non-overlapping, the two properties that make case totals safe to add. Tool #2 (Make a Systematic List) counts each case by walking through the rabbits in a fixed order, so no assignment is built twice. Tool #15 (Organize Information in More Ways) is held back for the final check, where the same count is rebuilt from the children's side instead of the parents'.

1STEP 1

Turn four bans into one rule

The bans reduce to one disjointness rule.

legal⇔ P∩ C=∅⇔every child avoids every store holding a parent
2STEP 2

Split on how many stores the parents fill

Only two cases can happen.

|P|∈{1,2}, #{ways to place the children}=(4-|P|)³
3STEP 3

Case 1: parents in one store

One place for the first group gives 108.

4 · 3³=4 · 27=108
4STEP 4

Case 2: parents in two stores

Two places gives 96.

4·3·2³=12·8=96
5STEP 5

Add the two cases

Adding gives 204, choice (D).

108+96=204 → (D)
Answer
204
Size check first: with no rule at all there are 4⁵=1024 assignments, so 204 is about a fifth of them, which is sensible for a ban that closes one or two of four stores to the three children. The answer choices also mark the two ways to stop early, and both are visible in the work: 108 is choice (B) and is only the parents-together case, while 96 is choice (A) and is only the parents-apart case. Choice (C) 156 is the trap for treating Peter and Pauline as interchangeable, which turns the apart case into C(4, 2)·2³=6·8=48 instead of 12·8=96 and gives 48+108=156; the problem names both parents, so the ordered count 12 is the correct one. Since 204 is the sum of two disjoint, exhaustive cases and matches an independent recount below, choice (D) stands.
💡Key takeaway

The rule only ever closes the stores the parents are in, so settle the parents first, count how many stores that blocks, and let each child pick freely from whatever is left.

  • Turn four bans into one rule
  • Split on how many stores the parents fill
  • Case 1: parents in one store
  • Case 2: parents in two stores
  • Add the two cases

More like this

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