AMC 8 · 2020 · #10
Grade 7 countingProblem
Zara has a collection of 4 marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?
Pick an answer.
AMC 8 2020 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: Zara has $4$ distinct marbles — Aggie (A), Bumblebee (B), Steelie (S), Tiger (T) — and lines them up on a shelf. She wants to count the line-ups in which Steelie and Tiger are NOT next to each other. How many such arrangements are there?
Givens: $4$ distinct marbles: Aggie, Bumblebee, Steelie, Tiger; All $4$ marbles are placed in a row (each used exactly once); Steelie and Tiger must NOT be adjacent; Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $18$, (E) $24$
Unknowns: The number of arrangements of the $4$ marbles in a row in which Steelie and Tiger are NOT next to each other
Understand
Restated: Zara has $4$ distinct marbles — Aggie (A), Bumblebee (B), Steelie (S), Tiger (T) — and lines them up on a shelf. She wants to count the line-ups in which Steelie and Tiger are NOT next to each other. How many such arrangements are there?
Givens: $4$ distinct marbles: Aggie, Bumblebee, Steelie, Tiger; All $4$ marbles are placed in a row (each used exactly once); Steelie and Tiger must NOT be adjacent; Answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $18$, (E) $24$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The word 'not next to' is the classic trigger for Tool #16 (Complement). Counting arrangements where S and T avoid each other directly forces messy case-splitting by where S goes; counting the OPPOSITE event (S and T together) is much easier because we can glue them into a single block. We'll use Tool #2 (Systematic List) as a sanity check for the small total of $24$ arrangements, and Tool #3 (Eliminate) to confirm the answer matches one of the given choices.
Execute — Answer: C
7.SP.C.8 Step 1 - Count the total arrangements of the $4$ marbles with no restriction.
- Position $1$ has $4$ choices, position $2$ has $3$ left, position $3$ has $2$ left, position $4$ has $1$ left — the Fundamental Counting Principle.
💡 Grade 7 'compound events via organized lists' covers using the multiplication rule to count ordered arrangements of distinct objects.
7.SP.C.8 Step 2 - Now count the FORBIDDEN arrangements where Steelie and Tiger are next to each other.
- Glue Steelie and Tiger into a single block [ST].
- We now arrange $3$ items in a row: the block, Aggie, and Bumblebee.
- That gives $3 \times 2 \times 1 = 6$ orders.
💡 Treating the adjacent pair as one object is the standard 'block trick' for counting arrangements with an adjacency constraint.
4.OA.A.3 Step 3 - Inside the block, Steelie and Tiger can stand as ST or TS — $2$ internal orders.
- Multiply the block orders by the internal orders to get every forbidden line-up.
💡 Grade 4 multi-step word-problem multiplication: orderings outside the block times orderings inside the block.
4.OA.A.3 Step 4 Subtract the forbidden arrangements from the total to get the allowed ones (Steelie NOT next to Tiger).
💡 The complement rule: (what we want) = (everything) $-$ (what we don't want).
4.OA.A.3 Step 5 - Match $12$ to the answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $18$, (E) $24$.
- Only choice (C) equals $12$, so the answer is (C).
💡 Tool #3 finishes the multiple-choice question by eliminating every non-matching option.
7.SP.C.8 Count the total arrangements of the $4$ marbles with no restriction. Position $1 7.SP.C.8 Now count the FORBIDDEN arrangements where Steelie and Tiger are next to each ot 4.OA.A.3 Inside the block, Steelie and Tiger can stand as ST or TS — $2$ internal orders. 4.OA.A.3 Subtract the forbidden arrangements from the total to get the allowed ones (Stee 4.OA.A.3 Match $12$ to the answer choices: (A) $6$, (B) $8$, (C) $12$, (D) $18$, (E) $24$ Review
Reasonableness: The total $24$ is exactly $4!$, which is correct for $4$ distinct objects in a row. The forbidden count $12$ is half of $24$, which makes intuitive sense: among $4$ random items, the probability that two specific items are adjacent is $\tfrac{2 \cdot 3!}{4!} = \tfrac{12}{24} = \tfrac{1}{2}$, so 'not adjacent' is also $\tfrac{1}{2}$, giving $12$ good arrangements. The answer $12$ is in the middle of the choices ($6, 8, 12, 18, 24$), which is the expected magnitude for a balanced complement count.
Alternative: Tool #2 (Systematic List) directly: with $24$ total arrangements, list them in dictionary order and cross out any with S adjacent to T. For example A B S T (bad: S-T adjacent), A B T S (bad), A S B T (good), A S T B (bad), A T B S (good), A T S B (bad), A S T B etc. Carrying this through all $24$ rows yields exactly $12$ survivors — same answer (C). Slower than the complement, but builds intuition for why exactly half the arrangements break the rule.
CCSS standards used (min grade 7)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying the $6$ block orders by the $2$ internal orders to get $12$ forbidden arrangements, then subtracting from the total $24$ to get $12$ allowed arrangements.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $4! = 24$ total arrangements and the $3! = 6$ block orders using the Fundamental Counting Principle, which lives inside Grade 7 compound-event counting.)
⭐ This AMC 8 problem only needs Grade 7 counting with organized lists you already know — count everything, count the bad cases, then subtract!
⭐ This AMC 8 problem only needs Grade 7 counting with organized lists you already know — count everything, count the bad cases, then subtract!
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