AMC 8 · 2008 · #14
Grade 7 countingProblem
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Pick an answer.
AMC 8 2008 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: Three $\text{A}$s, three $\text{B}$s, and three $\text{C}$s are placed in a $3 \times 3$ grid so that every row and every column contains exactly one of each letter. An $\text{A}$ is already placed in the upper-left corner. How many different ways can the rest of the grid be filled?
Givens: A $3 \times 3$ grid has $9$ cells; Three $\text{A}$s, three $\text{B}$s, three $\text{C}$s are used (nine letters total); Each row contains one $\text{A}$, one $\text{B}$, and one $\text{C}$; Each column contains one $\text{A}$, one $\text{B}$, and one $\text{C}$; An $\text{A}$ is fixed at position $(1,1)$, the upper-left corner; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Unknowns: The number of valid arrangements of the remaining eight cells
Understand
Restated: Three $\text{A}$s, three $\text{B}$s, and three $\text{C}$s are placed in a $3 \times 3$ grid so that every row and every column contains exactly one of each letter. An $\text{A}$ is already placed in the upper-left corner. How many different ways can the rest of the grid be filled?
Givens: A $3 \times 3$ grid has $9$ cells; Three $\text{A}$s, three $\text{B}$s, three $\text{C}$s are used (nine letters total); Each row contains one $\text{A}$, one $\text{B}$, and one $\text{C}$; Each column contains one $\text{A}$, one $\text{B}$, and one $\text{C}$; An $\text{A}$ is fixed at position $(1,1)$, the upper-left corner; Answer choices: (A) $2$, (B) $3$, (C) $4$, (D) $5$, (E) $6$
Plan
Primary tool: #13 Count Carefully
Secondary: #7 Break into Subproblems, #1 Draw a Picture
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 $\text{A}$s go first, then the $\text{B}$s, then the $\text{C}$s. Tool #1 (Draw a Picture) keeps the $3 \times 3$ 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.
Execute — Answer: C
7.SP.C.8 Step 1 - Draw the grid and place the rest of the $\text{A}$s.
- The given $\text{A}$ sits at $(1,1)$, so row $1$ and column $1$ already have their $\text{A}$.
- The other two $\text{A}$s must lie inside the $2 \times 2$ block formed by rows $2,3$ and columns $2,3$, with one $\text{A}$ per row and one per column.
- That is a $2 \times 2$ Latin choice: either the diagonal pair $(2,2)$ and $(3,3)$, or the anti-diagonal pair $(2,3)$ and $(3,2)$.
💡 Drawing the grid makes it clear that the $\text{A}$ in $(1,1)$ blocks row $1$ and column $1$, leaving a $2 \times 2$ subgrid where the other two $\text{A}$s must form a diagonal or anti-diagonal pair.
7.SP.C.8 Step 2 - Place the $\text{B}$s next.
- Fix one of the two $\text{A}$ patterns, say $\text{A}$s on the main diagonal: $(1,1), (2,2), (3,3)$.
- Row $1$ still needs a $\text{B}$ in cell $(1,2)$ or $(1,3)$.
- Whichever cell you pick for that $\text{B}$, every other $\text{B}$ position is forced by the row/column rule.
- Trying both: putting $\text{B}$ at $(1,2)$ forces $\text{B}$s at $(2,3)$ and $(3,1)$; putting $\text{B}$ at $(1,3)$ forces $\text{B}$s at $(2,1)$ and $(3,2)$.
- So once the $\text{A}$ pattern is fixed, there are $2$ valid $\text{B}$ patterns.
💡 Breaking the placement into A-then-B turns a tangled grid problem into two small choices in a row. The first $\text{B}$ in row $1$ determines everything else.
7.SP.C.8 Step 3 - Place the $\text{C}$s.
- After the $\text{A}$s and $\text{B}$s sit in five of the nine cells (the corner $\text{A}$, two more $\text{A}$s, three $\text{B}$s), each row and each column already has one $\text{A}$ and one $\text{B}$.
- The remaining three empty cells, one per row and one per column, must each get a $\text{C}$ — no choice involved.
💡 Two of the three letters in each row and column are settled, so the third letter has only one spot left. The $\text{C}$s are forced.
7.SP.C.8 Step 4 - Combine the counts using the multiplication principle (Tool #13).
- The three subproblems are independent in sequence: any $\text{A}$ pattern can be paired with either $\text{B}$ pattern, and the $\text{C}$ pattern is then forced.
💡 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.8 Draw the grid and place the rest of the $\text{A}$s. The given $\text{A}$ sits a 7.SP.C.8 Place the $\text{B}$s next. Fix one of the two $\text{A}$ patterns, say $\text{A 7.SP.C.8 Place the $\text{C}$s. After the $\text{A}$s and $\text{B}$s sit in five of the 7.SP.C.8 Combine the counts using the multiplication principle (Tool #13). The three subp Review
Reasonableness: List the four valid grids explicitly. With $\text{A}$s on the main diagonal (cells $(1,1),(2,2),(3,3)$) the two grids are rows $(\text{A},\text{B},\text{C}),(\text{C},\text{A},\text{B}),(\text{B},\text{C},\text{A})$ and rows $(\text{A},\text{C},\text{B}),(\text{B},\text{A},\text{C}),(\text{C},\text{B},\text{A})$. With $\text{A}$s on the anti-diagonal of the bottom $2\times 2$ block (cells $(1,1),(2,3),(3,2)$) the two grids are rows $(\text{A},\text{B},\text{C}),(\text{B},\text{C},\text{A}),(\text{C},\text{A},\text{B})$ and rows $(\text{A},\text{C},\text{B}),(\text{C},\text{B},\text{A}),(\text{B},\text{A},\text{C})$. That is exactly $4$ grids, all distinct. Cross-check: the total number of $3 \times 3$ Latin squares on three symbols is $12$, and by symmetry exactly $\tfrac{12}{3} = 4$ of them have an $\text{A}$ in the upper-left. Both checks agree.
Alternative: Tool #3 (Set Up an Equation) / direct enumeration: write the first row as $\text{A}\,\square\,\square$ where the two squares are some order of $\{\text{B},\text{C}\}$ — $2$ choices. Then write the first column as $\text{A}\,\square\,\square$ where the two squares are some order of $\{\text{B},\text{C}\}$ — another $2$ choices. After these four cells are filled, each remaining cell has exactly one letter that fits both its row and column, so the grid completes uniquely. Total: $2 \times 2 = 4$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation (Counting arrangements by breaking the placement into independent stages (A-pattern, then B-pattern, then forced C-pattern) and multiplying the choices, the Grade 7 compound-event counting rule.)
⭐ Place the letters one type at a time: $2$ ways for the rest of the $\text{A}$s, $2$ ways for the $\text{B}$s, then the $\text{C}$s have no choice — $2 \times 2 \times 1 = 4$ grids.
⭐ Place the letters one type at a time: $2$ ways for the rest of the $\text{A}$s, $2$ ways for the $\text{B}$s, then the $\text{C}$s have no choice — $2 \times 2 \times 1 = 4$ grids.
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