Problem 14 · 2022 AMC 8
Medium
Counting & Probability
careful-counting
In how many ways can the letters in BEEKEEPER be rearranged so that two or more E's do not appear together?
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Answer: D — 24 ways.
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Hint 1 of 2
BEEKEEPER is 5 E's and just 4 other letters in 9 slots. With so many E's crammed into 9 spots and none allowed to touch, the E placement is almost forced — figure out where they have to go.
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Hint 2 of 2
To keep 5 E's all apart in a row of 9, picture them with a non-E in each gap between them: E _ E _ E _ E _ E. That uses all 9 slots one way only. Then the 4 leftover letters fill the gaps.
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Approach: place the crowded letter first — it's forced — then permute the rest
- Insight: the 5 E's are the constraint, so place them first. To keep all 5 apart you need a non-E between every pair: E _ E _ E _ E _ E. That's 5 E's plus 4 separators = 9 slots — it fits with zero room to spare, so the E's must sit in positions 1, 3, 5, 7, 9.
- That leaves positions 2, 4, 6, 8 for B, K, P, R in any order: 4! = 24 arrangements.
- You'll see this again: in “keep these apart” problems, set down the troublesome items first as a frame (gaps between them), then slot the others into the gaps. When the items barely fit, their pattern is forced.
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