AMC 8 · 2008 · #14
Grade 7 countingPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a counting problem with two grid rules, so Tool #13 (Count Carefully) drives the work. To avoid mistakes we use Tool #7 (Break into Subproblems): decide where the rest of the As go first, then the Bs, then the Cs. Tool #1 (Draw a Picture) keeps the 3 × 3 grid in front of us so we can see what is fixed and what is free at each step. The multiplication principle then turns the three subproblem counts into one product.
The corner A fills row 1 and column 1, so the other two As sit in the bottom-right 2 × 2 block as a diagonal or anti-diagonal pair — 2 ways.
Drawing the grid makes it clear that the A in (1,1) blocks row 1 and column 1, leaving a 2 × 2 subgrid where the other two As must form a diagonal or anti-diagonal pair.
7.SP.C.8Draw A DiagramFix the A pattern; the top row still needs a B in one of two cells, and that single choice forces every other B — 2 ways.
Breaking the placement into A-then-B turns a tangled grid problem into two small choices in a row. The first B in row 1 determines everything else.
7.SP.C.8Identify SubproblemsWith an A and a B already in every row and column, the three empty cells each take the only letter left — 1 way for the Cs.
Two of the three letters in each row and column are settled, so the third letter has only one spot left. The Cs are forced.
7.SP.C.8Identify SubproblemsMultiply the independent counts with the multiplication principle: 2 × 2 × 1 gives 4 grids in all.
Multiplying the choice counts is the Grade 7 "compound event" rule. Two A-patterns times two B-patterns times one forced C-pattern equals 4.
7.SP.C.8Convert To AlgebraPlace the letters one type at a time: 2 ways for the rest of the As, 2 ways for the Bs, then the Cs have no choice — 2 × 2 × 1 = 4 grids.