AMC 10 · 2007 · #20
Grade 7 geometry-2dA set of 25 square blocks is arranged into a 5×5 square. How many different combinations of 3 blocks can be selected from that set so that no two are in the same row or column?
Pick an answer.
AMC 10 2007 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: In a $5 \times 5$ grid of $25$ square blocks, count how many ways there are to pick a group of $3$ blocks so that no two picked blocks share a row or share a column.
Givens: The blocks form a $5 \times 5$ square, so there are $5$ rows and $5$ columns; Exactly $3$ blocks are selected; No two selected blocks may lie in the same row or the same column; Answer choices: (A) $100$, (B) $125$, (C) $600$, (D) $2300$, (E) $3600$
Unknowns: The number of different $3$-block groups meeting the no-shared-row and no-shared-column rule
Understand
Restated: In a $5 \times 5$ grid of $25$ square blocks, count how many ways there are to pick a group of $3$ blocks so that no two picked blocks share a row or share a column.
Givens: The blocks form a $5 \times 5$ square, so there are $5$ rows and $5$ columns; Exactly $3$ blocks are selected; No two selected blocks may lie in the same row or the same column; Answer choices: (A) $100$, (B) $125$, (C) $600$, (D) $2300$, (E) $3600$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #16 Change Focus / Count the Complement
The rule "no two blocks share a row or column" means the $3$ chosen blocks sit in $3$ distinct rows and $3$ distinct columns. That lets Tool #7 (Identify Subproblems) split the count into three independent decisions: (1) which $3$ of the $5$ rows to use, (2) which $3$ of the $5$ columns to use, and (3) how to pair those columns with those rows. Tool #2 (Make a Systematic List) handles each small count, and Tool #16 (Change Focus) is the key reframe — instead of hunting for legal block-triples directly, we choose rows, choose columns, then match them.
Execute — Answer: C
7.SP.C.8 Step 1 Reframe as rows, columns, matching
- A block is named by its (row, column).
- If no two blocks share a row and no two share a column, then the $3$ blocks live in $3$ different rows and $3$ different columns.
- So choosing the blocks is the same as three separate jobs: choose which $3$ rows, choose which $3$ columns, then decide which chosen column goes with which chosen row.
💡 Each block sits at a crossing of one row and one column, so picking the blocks is really picking the rows, the columns, and how they cross.
7.SP.C.8 Step 2 Count ways to pick the rows
- Choose which $3$ of the $5$ rows will be used.
- List them by which two rows are left out: leaving out $\{1,2\},\{1,3\},\ldots,\{4,5\}$ gives every unordered choice exactly once.
- There are $10$ such pairs, so there are $10$ ways to pick the $3$ rows.
- The columns work identically: another $10$ ways.
💡 Choosing 3 to keep out of 5 is the same as choosing 2 to leave out, and there are only 10 ways to leave out a pair.
7.SP.C.8 Step 3 Count ways to match them up
- Now the $3$ rows and $3$ columns are fixed.
- Line up the $3$ chosen columns and assign each one a row from the $3$ chosen rows.
- The first column can take any of the $3$ rows, the second column any of the $2$ remaining rows, and the last column is forced.
- That is $3 \times 2 \times 1 = 6$ matchings, each giving a different legal set of $3$ blocks.
💡 Handing out 3 different rows to 3 different columns is just arranging 3 things in order: 6 ways.
4.OA.A.3 Step 4 Multiply the three counts
- The three choices are independent, so multiply them: $10$ ways for the rows, times $10$ ways for the columns, times $6$ ways to match.
- This counts every valid $3$-block group exactly once.
💡 When one choice does not affect another, the total number of combinations is the product of the separate counts.
7.SP.C.8 A block is named by its (row, column). If no two blocks share a row and no two s 7.SP.C.8 Choose which $3$ of the $5$ rows will be used. List them by which two rows are l 7.SP.C.8 Now the $3$ rows and $3$ columns are fixed. Line up the $3$ chosen columns and a 4.OA.A.3 The three choices are independent, so multiply them: $10$ ways for the rows, tim Review
Reasonableness: The answer $600$ must beat the plain (rule-free) count of $\binom{25}{3} = 2300$ picks of $3$ blocks, and it does — the row/column rule throws many groups out, so a number well below $2300$ is expected, ruling out (D) $2300$ and (E) $3600$. A second, independent count confirms $600$: place the blocks one at a time. The first block has $25$ choices; it blocks its whole row and column, leaving $16$ legal cells for the second; the second block leaves $9$ legal cells for the third. That is $25 \times 16 \times 9 = 3600$ ordered ways, but the group is unordered, so divide by $3! = 6$: $3600 / 6 = 600$. Both methods give $600$, matching (C).
Alternative: Tool #16 (Change Focus) with sequential placement, shown in the check above: $25 \times 16 \times 9$ ordered placements divided by $3! = 6$ for order gives $600$ directly, without ever splitting rows from columns.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Reframing the block selection as a compound choice (rows, then columns, then a matching) and using organized lists to count $\binom{5}{3}=10$ row choices, $10$ column choices, and $3!=6$ matchings.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the three independent counts by multiplication, $10 \times 10 \times 6 = 600$, and cross-checking with $25 \times 16 \times 9 \div 6$.)
⭐ When picks can't share a row or column, choose the rows, choose the columns, then count how to line them up — and multiply the three counts together.
⭐ When picks can't share a row or column, choose the rows, choose the columns, then count how to line them up — and multiply the three counts together.
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