Problem 10 · 2020 AMC 8
Easy
Counting & Probability
complementary-counting
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?
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Answer: C — 12 ways.
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Hint 1 of 2
“Not next to each other” is awkward to count head-on. Flip it: count everything, then subtract the arrangements where they are next to each other.
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Hint 2 of 2
To count the bad ones, glue Steelie and Tiger into a single block (they're forced together): 3 items arrange 3! = 6 ways, and the block flips 2 ways → 12 bad. Subtract from 4! = 24.
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Approach: complementary counting with the glue trick
- All arrangements of 4 distinct marbles: 4! = 24.
- Bad ones (Steelie–Tiger touching): glue them into one block, leaving 3 items → 3! = 6 orders; the block itself is ST or TS → ×2 = 12 bad.
- Good = 24 − 12 = 12.
- You'll see this again as: for a “must be adjacent” condition, glue them into one unit; for “must not be adjacent,” count the total and subtract the glued case. Two tools that pair up on almost every seating problem.
Another way — gap method (place the others first):
- Place the Aggie and Bumblebee first: 2! = 2 orders. They create 3 gaps: _ A _ B _.
- Drop Steelie and Tiger into two different gaps so they can't touch: choose 2 of the 3 gaps and order them, 3 × 2 = 6 ways.
- 2 × 6 = 12. The gap method builds only the good arrangements directly — no subtracting.
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