AMC 10 · 2012 · #11

Grade 6 arithmetic
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
permutations-basic casework ↑ Prerequisites: permutations-basic
📏 Short solution 💡 2 insights
Problem
A dessert chef plans one dessert for each of the seven days from Sunday through Saturday. Each day's dessert is cake, pie, ice cream, or pudding, and the same dessert cannot appear two days in a row. Friday must be cake because of a birthday. Count how many different seven-day menus are possible.

Pick an answer.

(A)
729
(B)
972
(C)
1024
(D)
2187
(E)
2304

AMC 10 2012 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) drives the count: decide the desserts in a smart order instead of guessing. Tool #7 (Identify Subproblems) turns one hard "count the whole week" into seven easy "count one day" pieces that multiply together. Tool #5 (Look for a Pattern) spots that every day except Friday has the same 3 choices, which collapses the product into a single power of 3 and guards against the trap of also giving Friday a factor.

1STEP 1

Anchor Friday, then count outward

Friday is forced to be cake, so fix it and pick each day beside one already chosen: it need only differ from that neighbor, 4-1=3 ways.

4-1=3 choices for each non-Friday day
2STEP 2

Multiply the independent day-counts

The 6 non-Friday days each have 3 independent choices and Friday contributes 1, so the multiplication principle gives 3⁶.

3×3×3×3×3×3₆ days × 1 = 3⁶
3STEP 3

Evaluate the power of 3

Multiply step by step: 3²=9, 3³=27, 3⁴=81, 3⁵=243, and 3⁶=729, which is choice (A).

3⁶=729 → (A)
Answer
729
Check the biggest trap: if you forget that Friday is fixed and give all 7 days 3 choices, you get 3⁷=2187, which is exactly the distractor (D). Anchoring Friday removes one factor of 3, dropping 2187 to 3⁶=729 — so (A) is right and (D) is the intended mistake. Also 4⁷ (ignoring the no-repeat rule) would be far larger than any choice, confirming the constraints really do the pruning.
💡Key takeaway

Pin down the forced day first, then every other day has just 3 choices, so multiply: 3⁶=729.

  • Anchor Friday, then count outward
  • Multiply the independent day-counts
  • Evaluate the power of 3

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