Sensim Math Original · sm-12
SM Original Grade 3 countingProblem
A modern art museum lays out its galleries in a 4×4 grid of 16 identical display rooms. After hours, two surveillance drones — Alpha and Beta — each hover above exactly one room. Whenever a drone occupies a room, its cameras watch every room one step away horizontally, vertically, or diagonally (so a drone in a fully interior room watches all 8 neighboring rooms).
For the night-shift schedule to be valid, the two drones must hover above two different rooms, and neither drone may watch the room the other one is in. Because Alpha and Beta carry different sensor packages, swapping which drone occupies which room counts as a separate schedule.
How many valid (Alpha, Beta) schedules are there?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: On a $4 \times 4$ grid of $16$ museum rooms, place two distinguishable drones Alpha and Beta — one drone per room, two different rooms — so that neither drone is in a room watched by the other. "Watched" means horizontally, vertically, or diagonally one step away. Count the number of ordered (Alpha-room, Beta-room) schedules that satisfy this.
Givens: Grid: $4 \times 4 = 16$ display rooms; Each drone watches every room one step away (up to $8$ rooms, like a chess-king move); Alpha and Beta carry different sensors, so the assignment is ordered (Alpha at room $A$, Beta at room $B$ differs from the swap); The two drones occupy different rooms, and neither watches the other's room; Answer choices: (A) 128, (B) 144, (C) 152, (D) 156, (E) 168
Unknowns: The number of valid ordered (Alpha-room, Beta-room) placements
Understand
Restated: On a $4 \times 4$ grid of $16$ museum rooms, place two distinguishable drones Alpha and Beta — one drone per room, two different rooms — so that neither drone is in a room watched by the other. "Watched" means horizontally, vertically, or diagonally one step away. Count the number of ordered (Alpha-room, Beta-room) schedules that satisfy this.
Givens: Grid: $4 \times 4 = 16$ display rooms; Each drone watches every room one step away (up to $8$ rooms, like a chess-king move); Alpha and Beta carry different sensors, so the assignment is ordered (Alpha at room $A$, Beta at room $B$ differs from the swap); The two drones occupy different rooms, and neither watches the other's room; Answer choices: (A) 128, (B) 144, (C) 152, (D) 156, (E) 168
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities
Everything sits on a tiny $4 \times 4$ picture, so first use Tool #1 — **draw the grid** and color the rooms by where they sit. The picture instantly shows the $16$ rooms come in just three flavors by location: **corner** (only $3$ neighboring rooms), **side** (along an edge but not a corner — $5$ neighbors), and **interior** (any of the four innermost rooms — $8$ neighbors). Tool #7 (identify subproblems) splits the big count into one subproblem per flavor — Alpha sits in a corner / on a side / in the interior — and inside each subproblem Tool #2 (systematic listing via the multiplication rule) counts Beta's legal rooms. Finally Tool #3 (eliminate possibilities) checks that the total matches one of (A)-(E). Visual + counting tools are plenty for a $4 \times 4$ board — no algebra needed.
Execute — Answer: D
K.G.A.1 Step 1 - Sketch the $4 \times 4$ grid and sort the $16$ rooms by location.
- The **4 corner rooms** sit at the four corners of the grid; the **8 side rooms** are the non-corner rooms along the four outer edges (2 per edge × 4 edges); the **4 interior rooms** are the inner $2 \times 2$ block in the middle.
- The three flavors partition the grid: $4 + 8 + 4 = 16$, every room classified exactly once.
💡 Sorting rooms by position words like "corner", "side", and "inside" is the Kindergarten position-vocabulary skill.
K.G.A.1 Step 2 - Read off how many rooms a drone watches from each flavor.
- A drone in a **corner** has only $3$ neighboring rooms inside the grid (the two adjacent corners are off-board, so just the right-of, below, and diagonal).
- A drone on a **side** has $5$ neighbors (three along its own edge plus two reaching one row inward).
- A drone in the **interior** has the full $8$ neighbors.
- The three watching-counts $3, 5, 8$ are the key numbers per flavor.
💡 Counting the up-to-$8$ neighbors around a single room is still a Kindergarten-level adjacency activity.
3.OA.A.1 Step 3 - Break the big problem ("total schedules") into **3 subproblems** by where Alpha sits: ① corner, ② side, ③ interior.
- For each subproblem the rooms Beta may occupy are $16 - 1 - (\text{watches})$: subtract Alpha's own room (1) and the rooms Alpha watches.
- By the multiplication rule the contribution of that subproblem is $(\text{Alpha rooms}) \times (\text{Beta rooms})$, and the three contributions get added because the three flavors of Alpha-rooms are disjoint.
💡 Reading "$A$ choices, each with $B$ choices, gives $A \times B$ total" is the Grade 3 product-of-equal-groups meaning of multiplication.
3.OA.C.7 Step 4 - **Subproblem ①: Alpha in a corner.** Alpha rooms: $4$.
- The corner drone blocks its own room plus the $3$ it watches, so $1 + 3 = 4$ rooms are off-limits to Beta.
- Beta picks from $16 - 4 = 12$ rooms.
- Contribution: $4 \times 12 = 48$.
💡 $4 \times 12 = 48$ is fluent multiplication within $100$, a Grade 3 standard.
3.OA.C.7 Step 5 - **Subproblem ②: Alpha on a side room.** Alpha rooms: $8$.
- The side drone blocks $1 + 5 = 6$ rooms, leaving Beta with $16 - 6 = 10$.
- Contribution: $8 \times 10 = 80$.
💡 $8 \times 10 = 80$ is a basic Grade 3 multiplication-within-$100$ fact.
3.OA.C.7 Step 6 - **Subproblem ③: Alpha in the interior.** Alpha rooms: $4$.
- The interior drone blocks $1 + 8 = 9$ rooms, leaving Beta with $16 - 9 = 7$.
- Contribution: $4 \times 7 = 28$.
💡 $4 \times 7 = 28$ is another Grade 3 multiplication fact.
3.NBT.A.2 Step 7 - Add the three subproblem contributions to get the total schedule count: $48 + 80 + 28 = 156$.
- The three Alpha-flavors partition the $16$ rooms with no overlap, so the addition rule applies directly.
💡 Adding three two-digit numbers like $48 + 80 + 28$ is squarely within Grade 3 add-within-$1000$ fluency.
2.NBT.A.4 Step 8 - Match $156$ against the answer choices.
- (A) $128$ is exactly the corner + side contributions $48 + 80$ (forgetting the interior subproblem).
- (B) $144$ and (C) $152$ are near-miss values that come from undercounting Beta-rooms in one flavor.
- (E) $168$ overshoots by $12$, the typical slip of counting interior watches as $6$ instead of $8$.
- Our total $156$ is exactly **(D)**.
💡 Choosing the matching three-digit number from a short list is a Grade 2 three-digit comparison skill.
K.G.A.1 Sketch the $4 \times 4$ grid and sort the $16$ rooms by location. The **4 corner K.G.A.1 Read off how many rooms a drone watches from each flavor. A drone in a **corner* 3.OA.A.1 Break the big problem ("total schedules") into **3 subproblems** by where Alpha 3.OA.C.7 **Subproblem ①: Alpha in a corner.** Alpha rooms: $4$. The corner drone blocks i 3.OA.C.7 **Subproblem ②: Alpha on a side room.** Alpha rooms: $8$. The side drone blocks 3.OA.C.7 **Subproblem ③: Alpha in the interior.** Alpha rooms: $4$. The interior drone bl 3.NBT.A.2 Add the three subproblem contributions to get the total schedule count: $48 + 80 2.NBT.A.4 Match $156$ against the answer choices. (A) $128$ is exactly the corner + side c Review
Reasonableness: Cross-check by counting the complement. Total ordered placements with Alpha and Beta in different rooms: $16 \times 15 = 240$. Now count ordered pairs of rooms that ARE adjacent. Horizontally or vertically adjacent unordered pairs on the $4 \times 4$ grid: $4 \cdot 3 + 4 \cdot 3 = 24$. Diagonally adjacent unordered pairs: $3 \cdot 3 \cdot 2 = 18$. Total adjacent unordered pairs: $24 + 18 = 42$; doubling for the ordered version gives $84$. Non-attacking ordered placements: $240 - 84 = 156$, matching our casework total exactly. Sanity-wise, the interior flavor giving the smallest contribution ($28$) makes sense because interior rooms watch the most other rooms.
Alternative: An alternative is Tool #16 (Change Focus / Count the Complement): compute the $240$ total ordered placements in different rooms, count the $84$ ordered adjacent pairs, and subtract to get $240 - 84 = 156$ (this is the cross-check above). For a young student, though, splitting Alpha's rooms into corner / side / interior and adding is more concrete, so the main solution uses #7 + #2.
CCSS standards used (min grade 3)
K.G.A.1Describe positions of objects using above, below, beside, in front of (Classifying the $16$ rooms into corner / side / interior flavors and counting each room's up-to-$8$ neighbors.)2.NBT.A.4Compare two three-digit numbers using symbols (Matching the computed total $156$ to the five three-digit answer choices to identify (D).)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Reading "(Alpha rooms) × (Beta rooms)" as the count per flavor — a product of equal-sized groups (the multiplication rule).)3.OA.C.7Fluently multiply and divide within 100 (Computing each flavor's contribution: $4 \times 12 = 48$, $8 \times 10 = 80$, and $4 \times 7 = 28$.)3.NBT.A.2Fluently add and subtract within 1000 (Adding the three flavor contributions $48 + 80 + 28 = 156$.)
⭐ This AMC 8 problem only needs Grade 3 multiplication and the "split into cases, then add" idea you already know!
⭐ This AMC 8 problem only needs Grade 3 multiplication and the "split into cases, then add" idea you already know!
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