AMC 8 · 2013 · #21
Grade 7 countingProblem
Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?
Pick an answer.
AMC 8 2013 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.
Toolkit + CCSS Solution
Understand
Restated: Samantha lives $2$ blocks west and $1$ block south of the southwest (SW) corner of City Park. Her school is $2$ blocks east and $2$ blocks north of the northeast (NE) corner of the park. She bikes the shortest street route to the SW corner, takes the single diagonal path across the park to the NE corner, then bikes the shortest street route to school. How many different shortest routes are possible in total?
Givens: Home is $2$ blocks W and $1$ block S of the park's SW corner; School is $2$ blocks E and $2$ blocks N of the park's NE corner; Inside the park: exactly $1$ diagonal path from SW corner to NE corner; On streets, she only moves along grid blocks (no diagonal shortcuts); Each leg must be a shortest path; Answer choices: (A) $3$, (B) $6$, (C) $9$, (D) $12$, (E) $18$
Unknowns: The total number of different shortest routes from home to school
Understand
Restated: Samantha lives $2$ blocks west and $1$ block south of the southwest (SW) corner of City Park. Her school is $2$ blocks east and $2$ blocks north of the northeast (NE) corner of the park. She bikes the shortest street route to the SW corner, takes the single diagonal path across the park to the NE corner, then bikes the shortest street route to school. How many different shortest routes are possible in total?
Givens: Home is $2$ blocks W and $1$ block S of the park's SW corner; School is $2$ blocks E and $2$ blocks N of the park's NE corner; Inside the park: exactly $1$ diagonal path from SW corner to NE corner; On streets, she only moves along grid blocks (no diagonal shortcuts); Each leg must be a shortest path; Answer choices: (A) $3$, (B) $6$, (C) $9$, (D) $12$, (E) $18$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #1 Draw a Diagram
The trip has three independent legs — Home $\to$ SW corner, SW $\to$ NE through the park, NE corner $\to$ School — so Tool #7 (Identify Subproblems) splits one hard count into three easy ones. Tool #2 (Systematic List) handles each street leg: with only $3$ or $4$ moves, we can list every shortest path by writing the move sequence (E and N letters) in order. Tool #1 (Draw a Diagram) keeps the directions straight on a grid so we never confuse N/S/E/W. Finally we multiply the three sub-counts (fundamental counting principle) because the legs are independent.
Execute — Answer: E
4.OA.A.3 Step 1 - Draw a grid.
- Place the park as a square; mark Home $2$ blocks W and $1$ block S of the SW corner, and School $2$ blocks E and $2$ blocks N of the NE corner.
- The shortest street route from Home to the SW corner uses only E (east) and N (north) moves: she needs $2$ E's and $1$ N.
- Any other direction would add length.
💡 Drawing the grid turns a confusing N/S/E/W word problem into a clear right-and-up walk — a Grade 4 multi-step word-problem setup.
7.SP.C.8 Step 2 - List every shortest path for Leg 1 as a sequence of $3$ letters made of $2$ E's and $1$ N.
- Use a fixed ordering rule: where does the N go (slot 1, 2, or 3)?
- That gives $3$ sequences: NEE, ENE, EEN.
- So Leg 1 has $\mathbf{3}$ shortest paths.
💡 An organized list of move sequences is exactly the Grade 7 "sample space" technique for counting compound events.
4.OA.A.3 Step 3 - Leg 2 is inside the park.
- The problem states she takes "a diagonal path" from the SW corner to the NE corner, and there is only one such diagonal.
- So Leg 2 contributes exactly $\mathbf{1}$ path.
💡 Splitting off the trivial leg early keeps the harder counting clean — the heart of Tool #7.
7.SP.C.8 Step 4 - For Leg 3 (NE corner $\to$ School), she needs $2$ E's and $2$ N's, total $4$ moves.
- List every shortest path as a sequence of $4$ letters with $2$ E's and $2$ N's.
- Order them by the position of the first N.
- The $6$ sequences are: NNEE, NENE, NEEN, ENNE, ENEN, EENN.
- So Leg 3 has $\mathbf{6}$ shortest paths.
💡 A systematic ordering rule guarantees we miss no sequence and double-count none — the discipline behind Tool #2.
7.SP.C.8 Step 5 The three legs are independent, so by the multiplication principle (fundamental counting principle), the total number of shortest routes is the product of the three counts.
💡 When choices in stage A do not affect choices in stage B, you multiply the counts — Grade 7 compound-event reasoning.
4.OA.A.3 Draw a grid. Place the park as a square; mark Home $2$ blocks W and $1$ block S 7.SP.C.8 List every shortest path for Leg 1 as a sequence of $3$ letters made of $2$ E's 4.OA.A.3 Leg 2 is inside the park. The problem states she takes "a diagonal path" from th 7.SP.C.8 For Leg 3 (NE corner $\to$ School), she needs $2$ E's and $2$ N's, total $4$ mov 7.SP.C.8 The three legs are independent, so by the multiplication principle (fundamental Review
Reasonableness: The answer $18$ matches choice (E). A sanity check: for an $m \times n$ grid path with only E and N moves, the count is $\binom{m+n}{m}$. Leg 1: $\binom{2+1}{1} = 3$. Leg 3: $\binom{2+2}{2} = 6$. Product $3 \times 1 \times 6 = 18$ — matches our enumeration exactly. The number is also in the expected range: with $3$ choices on the short leg and $6$ on the longer one, no answer below $6$ or above $18$ is plausible.
Alternative: Tool #9 (Solve an Easier Related Problem) gives the same answer. Start with the smallest grids: a $1 \times 1$ corner has $2$ shortest paths (EN, NE), a $1 \times 2$ corner has $3$, a $2 \times 2$ corner has $6$. The pattern matches Pascal's triangle ($\binom{m+n}{m}$), confirming Leg 1 = $3$ and Leg 3 = $6$ without re-listing. Multiply by $1$ for the fixed diagonal to get $18$.
CCSS standards used (min grade 7)
4.OA.A.3Solve multistep word problems with the four operations (Setting up the grid from the N/S/E/W description and splitting the trip into three independent legs to be combined at the end.)7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Enumerating the shortest paths on each street leg as ordered move sequences ($3$ for Leg 1, $6$ for Leg 3) and applying the multiplication principle to combine the three legs into $3 \times 1 \times 6 = 18$.)
⭐ Break the trip into pieces, list the few short paths in each piece, then multiply — that is all you need to count grid routes.
⭐ Break the trip into pieces, list the few short paths in each piece, then multiply — that is all you need to count grid routes.
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