AMC 8 · 2013 · #21

Grade 7 counting

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

lattice-pathscombinations-basicsystematic-enumeration identify-subproblemssystematic-enumeration ↑ Prerequisites: combinations-basicsystematic-enumeration

📏 Medium solution 💡 3 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The trip has three independent legs — Home → SW corner, SW → NE through the park, NE corner → 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.

1STEP 1

Draw the grid: reaching the SW corner from Home uses only East and North moves — 2 E and 1 N, 3 moves.

Leg 1 moves = 2E + 1N, total 3 moves

2STEP 2

Order the 2 E's and 1 N by where the N sits: NEE, ENE, EEN — so Leg 1 has 3 shortest paths.

{NEE, ENE, EEN} → 3 paths

3STEP 3

Inside the park only the single diagonal runs SW → NE, so Leg 2 gives exactly 1 path.

Leg 2 paths = 1

4STEP 4

Leg 3 (NE corner → School) needs 2 E's and 2 N's; ordering them gives 6 shortest paths.

{NNEE, NENE, NEEN, ENNE, ENEN, EENN} → 6 paths

5STEP 5

The legs are independent, so multiply the counts: 3 × 1 × 6 = 18 shortest routes.

Total = 3 × 1 × 6 = 18 → (E)

Answer

The answer 18 matches choice (E). A sanity check: for an m × n grid path with only E and N moves, the count is C(m+n, m). Leg 1: C(2+1, 1) = 3. Leg 3: C(2+2, 2) = 6. Product 3 × 1 × 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.

💡Key takeaway

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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