Competition · AMC preparation · step 4 of 4
AMC 8 · 2020 · #21
Grade 5 counting
Pick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Path-counting on a grid is the textbook job for Tool #1 (Draw a Diagram): redraw the board with P at the bottom and write the number of paths into each reachable white square, working one row at a time. Tool #5 (Look for a Pattern) shows up because the count at each square is the sum of the counts at the two diagonal squares below it — exactly the Pascal's-triangle pattern, modified at the edges where one diagonal falls off the board. Tool #7 (Identify Subproblems) makes the bookkeeping clean: "how many ways to reach (r,c)" is solved by adding the answers to the two smaller subproblems "how many ways to reach (r-1, c-1)" and "how many ways to reach (r-1, c+1)".
Set up the grid coordinates
Label columns 0-7 left-to-right and rows 0-7 bottom-to-top; from the figure P sits at (0, 5) and Q at (7, 6).
Setting up rows and columns on the board is the Grade 5 coordinate-grid idea: every white square gets an address (r, c).
5.G.A.1Draw A DiagramState the counting rule
Let N(r, c) = paths to square (r, c); a square is fed by its two lower diagonals, so N(r, c) = N(r-1, c-1) + N(r-1, c+1), N(0, 5) = 1.
Splitting "reach (r, c)" into the two smaller subproblems "reach the two diagonal neighbors below" is the Grade 4 "generate a pattern from a rule" move.
The number of paths that reach a white square equals the number of paths reaching its lower-left white neighbor plus the number reaching its lower-right white neighbor.
▸ Why?
Sort all paths that end on this square by which lower white neighbor the marker stepped from on its last move; every path lands in exactly one of these two groups and none is missed, so the total count is the two group counts added together.
▸ Why?
A step only rises diagonally from the row directly below, and just two white squares in that row touch this one, so no path can arrive from anywhere else and no single path can arrive from both — the two groups have no gaps and no overlaps.
▸ Why?
Each of those two group counts equals the number of paths that reach the matching neighbor, because a path arriving here through a neighbor is just a path to that neighbor with one fixed final step added on.
▸ Why?
Adding the same last step to each path that reaches the neighbor pairs it with exactly one path that reaches this square through that neighbor, so the two collections have equal size.
Fill in the first rows
Apply the rule upward - row 1: 1, 1; row 2: 1, 2, 1; row 3: 1, 3, 3 (column 8 is off-board, adding nothing there).
Each new number is the sum of the two diagonal numbers right below it — the same Pascal's-triangle pattern from Grade 4.
4.OA.C.5Look For A PatternContinue to the sixth row
Keep summing to row 6, edges inheriting one value - row 4: 4, 6, 3; row 5: 10, 9; row 6: 19, 9.
Writing the running total into each white square keeps the bookkeeping visual, with edge squares getting only one contribution.
4.OA.C.5Draw A DiagramAdd the two counts at Q
Finish at Q: N(7, 6) = 19 + 9 = 28, so there are 28 distinct 7-step paths from P to Q - choice (A).
Adding the two contributions 19 + 9 = 28 is a Grade 4 multi-digit addition.
4.NBT.B.4Draw A DiagramThis AMC 8 problem only needs Grade 5 coordinate grids and the Grade 4 "add the two numbers below" Pascal's-triangle pattern you already know!
- Set up the grid coordinates
- State the counting rule
- Fill in the first rows
- Continue to the sixth row
- Add the two counts at Q
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