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 six windows. In how many ways can
Georgie the Ghost enter the house by one window and leave
by a different window?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: The house has $6$ windows, each distinguishable; Georgie picks one window to enter and one window to exit; The exit window must be different from the entry window; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $30$, (E) $36$

Unknowns: The total number of distinct (entry window, exit window) pairs

Understand

Restated: 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?

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List

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 $6 \times 5$ gives the total.

Execute — Answer: D

#7 Identify Subproblems 3.OA.A.1 Step 1

Count the choices for the entry window.
Georgie can enter through any of the $6$ windows, so there are $6$ options for the first step.

$$\text{entry choices} = 6$$

💡 Pick one window from a set of $6$ — that is $6$ equal options for the first action.

#7 Identify Subproblems 3.OA.A.1 Step 2

Count the choices for the exit window.
Once the entry window is fixed, $1$ window is used up, so only $5$ windows remain available for exit.

$$\text{exit choices} = 6 - 1 = 5$$

💡 The "different window" rule removes exactly one window from the exit pool, leaving $5$.

#2 Make a Systematic List 4.OA.A.3 Step 3

Combine the two subproblems.
Every entry window pairs with every allowed exit window, so multiply the two counts to get the total number of trips.

$$\text{total} = 6 \times 5 = 30 \;\Rightarrow\; \textbf{(D)}$$

💡 A systematic list with $6$ entry rows, each followed by its $5$ allowed exits, has $6 \times 5 = 30$ rows total.

[1] #7 3.OA.A.1 Count the choices for the entry window. Georgie can enter through any of the $6$

[2] #7 3.OA.A.1 Count the choices for the exit window. Once the entry window is fixed, $1$ windo

[3] #2 4.OA.A.3 Combine the two subproblems. Every entry window pairs with every allowed exit wi

Review

Reasonableness: 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 $6 \times 5 = 30$. Also notice that $6 \times 6 = 36$ would count the impossible "same-window" trips; subtracting those $6$ leaves $36 - 6 = 30$, matching answer (D).

Alternative: Tool #3 (Eliminate Possibilities) on the choices: without the "different window" rule the count would be $6 \times 6 = 36$, which is choice (E). Removing the $6$ same-window trips drops the count by $6$, ruling out (E) and pointing to (D) $30 = 36 - 6$. Choices (A), (B), (C) are far too small to come from two events with $6$ and $5$ options.

CCSS standards used (min grade 4)

3.OA.A.1 Interpret products of whole numbers as combining groups of equal size (Reading $6 \times 5$ as "$6$ groups of $5$ exits", one group for each possible entry window.)
4.OA.A.3 Solve multistep word problems using the four operations (Combining the two counts ($6$ entry choices and $5$ exit choices) with one multiplication to get the total $30$.)

⭐ 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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