AMC 8 · 2020 · #10
Grade 7 countingPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
With no restriction, filling the 4 positions in order gives 4 × 3 × 2 × 1 = 24 total line-ups by the multiplication rule.
Grade 7 'compound events via organized lists' covers using the multiplication rule to count ordered arrangements of distinct objects.
7.SP.C.8Make A Systematic ListGlue Steelie and Tiger into one block [ST]; arranging that block with Aggie and Bumblebee gives 3 × 2 × 1 = 6 orders.
Treating the adjacent pair as one object is the standard 'block trick' for counting arrangements with an adjacency constraint.
7.SP.C.8Count The ComplementInside the block, ST or TS gives 2 internal orders, so forbidden line-ups total 6 × 2 = 12.
Grade 4 multi-step word-problem multiplication: orderings outside the block times orderings inside the block.
4.OA.A.3Count The ComplementSubtract the forbidden line-ups from the total to leave the ones with Steelie not beside Tiger: 24 - 12 = 12.
The complement rule: (what we want) = (everything) - (what we don't want).
4.OA.A.3Count The ComplementAmong the choices only (C) equals 12, so the answer is (C).
Tool #3 finishes the multiple-choice question by eliminating every non-matching option.
4.OA.A.3Eliminate PossibilitiesThis 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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