AMC 8 · 2024 · #13

Grade 2 counting

Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count

systematic-enumerationparity systematic-enumerationcomplementary-counting ↑ Prerequisites: multi-digit-arithmeticparity

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Buzz Bunny stands on the ground and makes a sequence of 6 hops; each hop is one stair up (U) or one stair down (D). Buzz must end back on the ground and can never go below the ground at any moment. We want to count how many different hop-sequences work.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The answer choices are tiny (the biggest is 12), so the entire universe of candidate sequences is small enough to LIST. Tool #2 (Systematic List) is exactly right: pick an ordering rule and write every candidate down once. To keep track of "is Buzz underground yet?" at each step, we use Tool #1 (Draw a Diagram) — a little stair-height number line under each sequence. Finally, Tool #3 (Eliminate Possibilities) crosses out any sequence that dips below 0, leaving only the valid ones to count. We deliberately avoid Tool #13 (Algebra) or the Catalan-number formula — an elementary student does not need them when listing is fast and trustworthy.

1STEP 1

Each U is +1 stair, each D is -1, and 6 hops must return to 0 — so the ups and downs balance: exactly 3 U's and 3 D's.

#U + #D = 6 and #U = #D → #U = #D = 3

2STEP 2

First hop must be U (else underground), last must be D (to land at 0), so every sequence is U _ _ _ _ D with 2 U's and 2 D's inside.

Sequence = U _ _ _ _ D with 2 U's and 2 D's in the middle

3STEP 3

Place the two middle U's by ordered slot-pairs (1,2),(1,3),(1,4),(2,3),(2,4),(3,4), giving 6 candidate sequences.

Candidates: UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD, UDDUUD

4STEP 4

Trace each height from 0 (+1 for U, -1 for D); all stay at or above 0 except UDDUUD, which hits -1 after three hops — eliminate it.

UDDUUD: 0 → 1 → 0 → -1 (below ground — invalid)

5STEP 5

Five sequences survive — UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD — so the answer is 5, choice (B).

Valid count = 6 - 1 = 5 → (B)

Answer

The answer 5 sits comfortably in the middle of the choices (4, 5, 6, 8, 12), and the structural facts agree with it: there are C(4, 2) = 6 ways to put 2 U's into 4 middle slots, and exactly one of those (UDDUUD) dips below ground, leaving 6 - 1 = 5. Each of the 5 surviving sequences is easy to verify by tracing the stair heights, and none of them are double-counted because the listing rule (positions of the two middle U's) is strict and increasing.

💡Key takeaway

This AMC 8 problem only needs Grade 2 step-by-step adding-and-subtracting you already know!

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