AMC 8 · 2007 · #4

Easy mode Grade 4

Problem

Picture a haunted house with six windows.

Georgie the Ghost wants to come in through one window and then leave through a window. The window he leaves by must be different from the window he came in by.

How many different ways can Georgie do this?

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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