AMC 8 · 2007 · #4

Grade 4 counting

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

permutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: multi-digit-arithmetic

📏 Short solution 💡 1 insight

Problem

A haunted house has 6 windows. Georgie the Ghost enters through one window and leaves through a different one. How many entry-exit pairs are possible?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The trip splits into two independent jobs: first pick the entry window, then pick the exit window. Tool #7 (Identify Subproblems) handles that split — count each piece on its own, then combine by multiplication. Tool #2 (Systematic List) is the safety net: imagine writing the entry window in a column and the exit window beside it, with the rule that the two must differ. Listing a few rows confirms each entry has exactly 5 matching exits, so 6 × 5 gives the total.

1STEP 1

Georgie can enter through any window, so the entry has 6 choices.

entry choices = 6

2STEP 2

The exit must differ from the entry, so one window is gone, leaving 5 exit choices.

exit choices = 6 - 1 = 5

3STEP 3

Each entry pairs with its 5 exits, so multiply: 6 × 5 = 30 trips.

total = 6 × 5 = 30 → (D)

Answer

Sanity check by listing a few rows. If Georgie enters through window 1, the exits are windows 2, 3, 4, 5, 6 — that is 5 trips. The same holds for entry windows 2, 3, 4, 5, 6, each giving 5 trips. Six rows of five gives 6 × 5 = 30. Also notice that 6 × 6 = 36 would count the impossible "same-window" trips; subtracting those 6 leaves 36 - 6 = 30, matching answer (D).

💡Key takeaway

Two actions in a row: count each, then multiply. Six entry windows times five remaining exit windows gives 30 — a Grade 4 multi-step multiplication problem.

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