Problem 13 · 2024 AMC 8
Medium
Counting & Probability
careful-countingcasework
Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of 6 hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)
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Answer: B — 5 sequences.
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Hint 1 of 2
Ending back on the ground after 6 hops forces exactly 3 ups and 3 downs. The real constraint: at no moment can downs outnumber ups, or Buzz drops below the ground.
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Hint 2 of 2
This is the "balanced parentheses" rule — read U as ( and D as ). List the legal arrangements; always start with U, end with D.
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Approach: count valid never-go-below sequences (a Catalan count)
- Back to the ground in 6 hops means equal ups and downs: 3 U and 3 D. The twist is the ground floor — reading left to right, the count of D's may never exceed the count of U's (else Buzz steps below the ground). So every valid string starts U and ends D.
- List them with that rule, keeping ups ahead: UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD — 5 in all.
- This transfers: "up/down steps that never go below start" is exactly the balanced-parentheses problem, and its counts are the Catalan numbers 1, 2, 5, 14, … For 3 ups and 3 downs the count is the 3rd Catalan number, 5 — matching our list, and a fast check for the longer versions of this problem.
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