AMC 8 · 2024 · #13

Grade 2 counting

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz Bunny 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.)

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Total number of hops: $6$; Each hop is either Up (+1 stair) or Down (-1 stair); Start position: ground (level $0$); End position: ground (level $0$); Buzz can never be below the ground (level must stay $\geq 0$ at every step); Five answer choices: (A) 4, (B) 5, (C) 6, (D) 8, (E) 12

Unknowns: The number of valid 6-hop sequences that start at the ground, end at the ground, and never go below it

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

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.

Execute — Answer: B

#1 Draw a Diagram 1.OA.A.1 Step 1

First nail down the structure.
Each U adds $1$ stair, each D takes away $1$ stair, and after all $6$ hops Buzz is back at $0$.
That means the total added must equal the total taken away, so the number of U's equals the number of D's.
With $6$ hops total, that forces exactly $3$ U's and $3$ D's.

$$\#U + \#D = 6 \;\text{and}\; \#U = \#D \;\Rightarrow\; \#U = \#D = 3$$

💡 Splitting $6$ hops into two equal groups of $3$ is a simple addition/subtraction word-problem move from Grade 1.

#1 Draw a Diagram 1.OA.A.1 Step 2

Two boundary observations shrink the list a lot.
The very first hop must be U, because a D would put Buzz at level $-1$ (underground).
The very last hop must be D, because to land at $0$ from a non-negative position the last move has to bring the level down.
So every valid sequence looks like $U \,\_\,\_\,\_\,\_\, D$, and the four blanks contain the remaining $2$ U's and $2$ D's.

$$\text{Sequence} = U\,\_\,\_\,\_\,\_\,D \;\text{with 2 U's and 2 D's in the middle}$$

💡 Reasoning "if I subtract $1$ from $0$ I go below zero, so the first move can't be a subtraction" is everyday Grade-1 add/subtract thinking.

#2 Make a Systematic List K.MD.B.3 Step 3

Now list every way to place $2$ U's and $2$ D's in the $4$ middle slots, using a systematic ordering rule: write the position of the two U's in increasing order — $(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)$.
That gives exactly $6$ candidate sequences.
Below I rebuild each full sequence by sandwiching the middle between the forced U and D.

$$\text{Candidates: } UUUDDD,\; UUDUDD,\; UUDDUD,\; UDUUDD,\; UDUDUD,\; UDDUUD$$

💡 Putting items in a fixed order so nothing is missed or repeated is the "classify and count" idea from Kindergarten.

#3 Eliminate Possibilities 2.OA.A.1 Step 4

For each of the $6$ candidates, draw the stair-height after each hop (start at $0$, $+1$ for U, $-1$ for D).
Cross out any sequence that ever drops below $0$.
The heights are: $UUUDDD\to (1,2,3,2,1,0)$ OK; $UUDUDD\to (1,2,1,2,1,0)$ OK; $UUDDUD\to (1,2,1,0,1,0)$ OK; $UDUUDD\to (1,0,1,2,1,0)$ OK; $UDUDUD\to (1,0,1,0,1,0)$ OK; $UDDUUD\to (1,0,-1,\ldots)$ \textbf{underground} — eliminate.

$$UDDUUD: 0 \xrightarrow{U} 1 \xrightarrow{D} 0 \xrightarrow{D} -1 \;\;\text{(below ground — invalid)}$$

💡 Tracking a running total with $+1$ and $-1$ across several steps and checking whether it stays at or above $0$ is a two-step add/subtract task from Grade 2.

#3 Eliminate Possibilities K.MD.B.3 Step 5

Count what survived: $UUUDDD,\; UUDUDD,\; UUDDUD,\; UDUUDD,\; UDUDUD$ — that is $5$ valid sequences.
So the answer is $\boxed{5}$, choice $\textbf{(B)}$.
Cross-checking the other options: (A) $4$ would mean we lost an extra valid sequence (we did not), (C) $6$ would mean we forgot to eliminate $UDDUUD$, and (D) $8$ and (E) $12$ overshoot even the total number of starts-with-U-ends-with-D candidates.

$$\text{Valid count} = 6 - 1 = 5 \;\Rightarrow\; \textbf{(B)}$$

💡 After we sorted the candidates into "valid" and "invalid" groups, counting how many are in the valid group is exactly the K-grade classify-and-count skill.

[1] #1 1.OA.A.1 First nail down the structure. Each U adds $1$ stair, each D takes away $1$ stai

[2] #1 1.OA.A.1 Two boundary observations shrink the list a lot. The very first hop must be U, b

[3] #2 K.MD.B.3 Now list every way to place $2$ U's and $2$ D's in the $4$ middle slots, using a

[4] #3 2.OA.A.1 For each of the $6$ candidates, draw the stair-height after each hop (start at $

[5] #3 K.MD.B.3 Count what survived: $UUUDDD,\; UUDUDD,\; UUDDUD,\; UDUUDD,\; UDUDUD$ — that is

Review

Reasonableness: 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 $\binom{4}{2} = 6$ 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.

Alternative: An alternative path is Tool #9 (Solve an Easier Related Problem): first do the problem with only $2$ hops (answer $1$: just $UD$), then $4$ hops (answer $2$: $UUDD$ and $UDUD$), then $6$ hops. The pattern $1, 2, 5, \ldots$ is the start of the Catalan numbers, which would predict $5$ for $6$ hops. This is a beautiful long-term observation, but the systematic-list approach above is faster and more honest for an elementary student because it shows exactly which sequences are valid.

CCSS standards used (min grade 2)

1.OA.A.1 Solve addition and subtraction word problems within 20 (Splitting $6$ hops into $3$ U's and $3$ D's, and reasoning that the first hop cannot be a subtraction from $0$.)
K.MD.B.3 Classify objects into given categories and count the numbers in each (Listing the $6$ candidate sequences in a strict order and then counting how many ended up in the "valid" category.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Tracking Buzz's stair height as a running total of $+1$ and $-1$ across the $6$ hops and checking it never falls below $0$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 2)

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