AMC 10 · 2012 · #8

Grade 6 counting
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
permutations-basic casework ↑ Prerequisites: permutations-basic
📏 Short solution 💡 2 insights
Problem
Each day's choice differs from the day before, and one day is fixed in advance. Count the possible weeks.

Pick an answer.

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

The fixed day is the anchor.

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

Multiply the independent day-counts

Every other day then has 3 options.

3×3×3×3×3×3₆ days × 1 = 3⁶
3STEP 3

Evaluate the power of 3

Multiplying gives 729, 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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