Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #21
Grade 7 countingPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Draw the route grid
Draw the grid: reaching the SW corner from Home uses only East and North moves — 2 E and 1 N, 3 moves.
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.
4.OA.A.3Draw A DiagramCount the first leg paths
Order the 2 E's and 1 N by where the N sits: NEE, ENE, EEN — so Leg 1 has 3 shortest paths.
An organized list of move sequences is exactly the Grade 7 "sample space" technique for counting compound events.
7.SP.C.8Make A Systematic ListCount the park path
Inside the park only the single diagonal runs SW → NE, so Leg 2 gives exactly 1 path.
Splitting off the trivial leg early keeps the harder counting clean — the heart of Tool #7.
4.OA.A.3Identify SubproblemsCount the third leg paths
Leg 3 (NE corner → School) needs 2 E's and 2 N's; ordering them gives 6 shortest paths.
A systematic ordering rule guarantees we miss no sequence and double-count none — the discipline behind Tool #2.
7.SP.C.8Make A Systematic ListMultiply the three counts
The legs are independent, so multiply the counts: 3 × 1 × 6 = 18 shortest routes.
When choices in stage A do not affect choices in stage B, you multiply the counts — Grade 7 compound-event reasoning.
The total number of different shortest routes equals the product of the three legs' path counts, 3 × 1 × 6.
▸ Why?
A full route is one home-to-park street path, then the single diagonal across the park, then one park-to-school street path, joined end to end — so the trip splits into three consecutive legs that share no streets and cover the whole journey with no gaps or overlaps.
▸ Why?
The choice on each leg is independent: which of the 3 home-to-park paths she takes does not limit which of the 6 park-to-school paths she can take, and the diagonal is forced — so pairing every independent choice with every other makes the count the product of the per-leg counts.
Break the trip into pieces, list the few short paths in each piece, then multiply — that is all you need to count grid routes.
- Draw the route grid
- Count the first leg paths
- Count the park path
- Count the third leg paths
- Multiply the three counts
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