Everyday at school, Jo climbs a flight of 6 stairs. Jo can take the stairs 1, 2, or 3 at a time. For example, Jo could climb 3, then 1, then 2. In how many ways can Jo climb the stairs?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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Toolkit + CCSS Solution
Understand
Restated: Jo climbs a $6$-stair flight, and each move covers $1$, $2$, or $3$ stairs. Counting the order in which the moves are made (so $3, 1, 2$ and $2, 1, 3$ are different), in how many different sequences can Jo reach the top?
Givens: Total of $6$ stairs to climb; Each step can be $1$, $2$, or $3$ stairs; Sequences are ordered: $1, 2$ and $2, 1$ count as two different ways; Answer choices: (A) $13$, (B) $18$, (C) $20$, (D) $22$, (E) $24$
Unknowns: The number of distinct ordered sequences of $1$s, $2$s, and $3$s whose terms sum to $6$
Understand
Restated: Jo climbs a $6$-stair flight, and each move covers $1$, $2$, or $3$ stairs. Counting the order in which the moves are made (so $3, 1, 2$ and $2, 1, 3$ are different), in how many different sequences can Jo reach the top?
Givens: Total of $6$ stairs to climb; Each step can be $1$, $2$, or $3$ stairs; Sequences are ordered: $1, 2$ and $2, 1$ count as two different ways; Answer choices: (A) $13$, (B) $18$, (C) $20$, (D) $22$, (E) $24$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem, #2 Make a Systematic List, #5 Look for a Pattern
Listing all sequences for $6$ stairs by hand is error-prone, so we use Tool #9 first: solve the easier versions $f(1)$, $f(2)$, $f(3)$ by Tool #2 (systematic listing). Then Tool #7 (Identify Subproblems) gives the key idea: whatever Jo's last move was — a $1$, a $2$, or a $3$ — the rest of the climb is just a smaller version of the same problem. That splits $f(n)$ into three subproblems: $f(n-1) + f(n-2) + f(n-3)$. Tool #5 (Pattern) then lets us extend the rule from the easy cases up to $n = 6$ without writing out a single full sequence.
Execute — Answer: E
#9 Solve an Easier Related Problem K.OA.A.1Step 1
Use Tool #9 to shrink the problem.
Start with $1$ stair: the only move is a single $1$.
So $f(1) = 1$.
$$f(1) = 1$$
💡 Solving the smallest case first is the Tool #9 move — it gives us a foothold.
#2 Make a Systematic List 1.OA.A.1Step 2
List the sequences for $2$ stairs using Tool #2: $(1,1)$ and $(2)$.
That's $f(2) = 2$.
Then for $3$ stairs: $(1,1,1)$, $(1,2)$, $(2,1)$, $(3)$ — so $f(3) = 4$.
$$f(2) = 2, \quad f(3) = 4$$
💡 Listing strictly by "length of the first move" keeps the count complete and duplicate-free.
#7 Identify Subproblems 4.OA.C.5Step 3
Now apply Tool #7.
For any $n \ge 4$, look at Jo's last move: it was either a $1$ (then the previous landing was stair $n-1$), a $2$ (previous was $n-2$), or a $3$ (previous was $n-3$).
Each case contributes $f(n-1)$, $f(n-2)$, $f(n-3)$ sequences.
Add them.
$$f(n) = f(n-1) + f(n-2) + f(n-3)$$
💡 Splitting by "what the last step was" turns one hard question into three easier ones — the heart of Tool #7.
#5 Look for a Pattern 4.OA.C.5Step 4
Use the rule from Step 3 to extend the pattern with Tool #5.
From $f(1) = 1$, $f(2) = 2$, $f(3) = 4$, compute $f(4)$, $f(5)$, $f(6)$ in turn.
$$f(4) = 4 + 2 + 1 = 7$$
💡 Each new term is just the sum of the three before it — a clean numeric pattern to extend.
💡 Reaching $f(6)$ took only two more additions — far faster than listing all $24$ sequences.
[1]
#9 K.OA.A.1Use Tool #9 to shrink the problem. Start with $1$ stair: the only move is a sing
[2]
#2 1.OA.A.1List the sequences for $2$ stairs using Tool #2: $(1,1)$ and $(2)$. That's $f(2)
[3]
#7 4.OA.C.5Now apply Tool #7. For any $n \ge 4$, look at Jo's last move: it was either a $1
[4]
#5 4.OA.C.5Use the rule from Step 3 to extend the pattern with Tool #5. From $f(1) = 1$, $f
[5]
#5 4.OA.C.5Keep the recurrence rolling to reach $f(6)$.
Review
Reasonableness: The answer $24$ is the largest choice, which fits intuition: a $6$-stair climb with three different step sizes gives many ordered combinations. We can spot-check $f(4) = 7$ by listing — $(1,1,1,1)$, $(1,1,2)$, $(1,2,1)$, $(2,1,1)$, $(2,2)$, $(1,3)$, $(3,1)$ — exactly $7$. The recurrence holds, so $f(6) = 24$ is consistent.
Alternative: Tool #2 (Systematic List) by partitions: the multisets of $1$s, $2$s, $3$s that sum to $6$ are $\{1,1,1,1,1,1\}$, $\{1,1,1,1,2\}$, $\{1,1,2,2\}$, $\{2,2,2\}$, $\{1,1,1,3\}$, $\{1,2,3\}$, $\{3,3\}$. The number of orderings of each is $1, 5, 6, 1, 4, 6, 1$ respectively. Sum: $1+5+6+1+4+6+1 = 24$. Same answer.
CCSS standards used (min grade 4)
K.OA.A.1 Represent addition and subtraction with objects, drawings, etc. (Establishing the base case $f(1) = 1$ — there is exactly one way to climb a single stair.)
1.OA.A.1 Use addition and subtraction within $20$ to solve word problems (Counting the small cases $f(2) = 2$ and $f(3) = 4$ by listing every valid sequence.)
4.OA.C.5 Generate and analyze patterns (Setting up the recurrence $f(n) = f(n-1) + f(n-2) + f(n-3)$ and extending it term by term to compute $f(4) = 7$, $f(5) = 13$, $f(6) = 24$.)
⭐ This AMC 8 problem only needs Grade 4 pattern reasoning — split by the last step and add the smaller cases — that you already know!
⭐ This AMC 8 problem only needs Grade 4 pattern reasoning — split by the last step and add the smaller cases — that you already know!
