AMC 10 · 2021 · #20
Grade 6 countingProblem
In how many ways can the sequence 1,2,3,4,5 be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
Pick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count the arrangements (permutations) of the sequence $1, 2, 3, 4, 5$ such that nowhere in the arrangement do three consecutive terms appear in strictly increasing order, and nowhere do three consecutive terms appear in strictly decreasing order.
Givens: Five distinct numbers: $1, 2, 3, 4, 5$; Forbidden: any three consecutive terms going up (e.g. $\dots a < b < c \dots$); Forbidden: any three consecutive terms going down (e.g. $\dots a > b > c \dots$); Choices: (A) $10$, (B) $18$, (C) $24$, (D) $32$, (E) $44$
Unknowns: Number of valid arrangements out of $5! = 120$
Understand
Restated: Count the arrangements (permutations) of the sequence $1, 2, 3, 4, 5$ such that nowhere in the arrangement do three consecutive terms appear in strictly increasing order, and nowhere do three consecutive terms appear in strictly decreasing order.
Givens: Five distinct numbers: $1, 2, 3, 4, 5$; Forbidden: any three consecutive terms going up (e.g. $\dots a < b < c \dots$); Forbidden: any three consecutive terms going down (e.g. $\dots a > b > c \dots$); Choices: (A) $10$, (B) $18$, (C) $24$, (D) $32$, (E) $44$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #2 Make a Systematic List, #5 Look for a Pattern, #16 Change Focus / Count the Complement
Tool #9 (Easier Problem) — solve for $n = 3$ and $n = 4$ first, where we can list everything by hand. Tool #2 (Systematic List) — enumerate alternating permutations in lexicographic order so nothing is missed. Tool #5 (Pattern) — the small-case counts $a_3 = 4, a_4 = 10$ already reveal the structure, and Tool #16 (symmetry between "up-down" and "down-up" via reversing inequalities) cuts the work in half. With those in hand, count $n = 5$ by symmetry plus a clean case-split on the position of $1$ (the unique smallest element).
Execute — Answer: D
6.NS.C.7 Step 1 - Reformulate: a triple $a < b < c$ or $a > b > c$ is forbidden.
- So for every interior position, the term there is either greater than both its neighbors (peak) or smaller than both (valley).
- That means signs of differences alternate — the arrangement zigzags up-down-up-down or down-up-down-up.
💡 Each interior term must be a peak or a valley — no "flat" run of three increasing or decreasing.
4.OA.B.4 Step 2 - Easier-Problem warmup: try $n = 3$.
- List all $3! = 6$ permutations of $\{1, 2, 3\}$: $123, 132, 213, 231, 312, 321$.
- Forbidden are the monotone ones $123$ and $321$.
- Valid: $132, 213, 231, 312$ — that is $4$ arrangements.
💡 Just drop the two monotone perms and count the rest.
5.OA.B.3 Step 3 - Easier-Problem step 2: try $n = 4$.
- Use Tool #16 (symmetry): valid arrangements split into those starting up ($a_1 < a_2$) and starting down ($a_1 > a_2$); reversing the inequalities pairs them, so the two halves are equal in count.
- Enumerate up-down-up arrangements of $\{1, 2, 3, 4\}$ systematically.
- The pattern is $a < b > c < d$.
- By listing: $1324, 1423, 2314, 2413, 3412$ — five up-down-up arrangements.
- Double by symmetry: $10$ total.
💡 Symmetry halves the work; then list up-down-up permutations by smallest first.
5.OA.B.3 Step 4 - Apply the same symmetry to $n = 5$: count arrangements of pattern $a < b > c < d > e$ (up-down-up-down) and double.
- Group by where the smallest element $1$ sits.
- Since $1$ is smaller than every neighbor, $1$ must be a valley — positions $1, 3$ or $5$.
- (Positions $2$ or $4$ would require $1$ to be a peak, impossible.)
💡 $1$ is the floor — it can only sit in valley positions.
5.OA.B.3 Step 5 - Sub-case (a): $1$ at position 1, pattern $1 < b > c < d > e$.
- The four remaining $\{2, 3, 4, 5\}$ fill $b, c, d, e$ with $b > c < d > e$.
- Tool #2 (Systematic List) by smallest of $\{c, e\}$: $c = 2$ gives $b > 2 < d > e$, so $\{b, d, e\} = \{3, 4, 5\}$ with $b > 2, d > 2, d > e$, i.e.
- any peak $d \in \{4, 5\}$ and the others split.
- $d = 5$: $\{b, e\} = \{3, 4\}$, $b, e$ free (no constraint between them), so $2$ orderings.
- $d = 4$: $\{b, e\} = \{3, 5\}$, $e < 4$ means $e = 3, b = 5$ — $1$ ordering.
- Subtotal $c = 2$: $3$.
- Symmetric: $e = 2$ gives $1 < b > c < d > 2$, by mirror-symmetry $3$ more arrangements.
- Together $6$ for sub-case (a).
💡 Pin $1$, then split by which interior valley equals $2$.
5.OA.B.3 Step 6 - Sub-case (b): $1$ at position 3, pattern $a < b > 1 < d > e$ with $\{a, b, d, e\} = \{2, 3, 4, 5\}$.
- Constraints: $a < b, d > e, b > 1$ (auto), $d > 1$ (auto).
- Free constraints: choose any $\{a, b\}$ as a $2$-subset of $\{2,3,4,5\}$ (with $a < b$ giving $\binom{4}{2} = 6$ ways) and the remaining two go to $\{d, e\}$ with $d > e$ ($1$ way).
- So $6$ arrangements.
💡 Split $\{2,3,4,5\}$ into left pair (ascending) and right pair (descending) — that's $\binom{4}{2}$ choices.
5.OA.B.3 Step 7 - Sub-case (c): $1$ at position 5, pattern $a < b > c < d > 1$.
- Mirror of sub-case (a) (read right-to-left): also $4$ arrangements?
- Let's recount directly.
- Remaining $\{a, b, c, d\} = \{2, 3, 4, 5\}$, constraints $a < b, b > c, c < d, d > 1$ (auto).
- Split by $c$: $c = 2$ gives $a < b > 2 < d$, $\{a, b, d\} = \{3, 4, 5\}$.
- Need $b > a$ and $b > 2$ (auto) and $d > 2$ (auto).
- $b \in \{3, 4, 5\}$, $a < b$, $\{a, d\} = \{3, 4, 5\} \setminus \{b\}$.
- $b = 5$: $\{a, d\} = \{3, 4\}$, $a$ can be $3$ or $4$ — but $a < 5$ auto, both work — $2$.
- $b = 4$: $\{a, d\} = \{3, 5\}$, need $a < 4$ so $a = 3, d = 5$ — $1$.
- $b = 3$: $a < 3$, $a \in \{4, 5\}$ impossible — $0$.
- Subtotal $c = 2$: $3$.
- By symmetric counting on the other end, $c = 3$ with $a, b > 3$ and one of $\{4, 5\}$...
- wait, let me re-split.
- Actually split by which of $\{c, e\}$...
- here $e$ is fixed as position with $1$.
- Just split by $c$: $c$ is a valley between $b$ and $d$, $c \in \{2, 3\}$ since $c < b, c < d$.
- $c = 3$: $a < b > 3 < d > 1$, $\{a, b, d\} = \{2, 4, 5\}$.
- Need $a < b$, $b > 3$, $d > 3$.
- $b, d \in \{4, 5\}$, so $\{b, d\} = \{4, 5\}$, $a = 2$.
- $b, d$ orderings: $2$.
- Subtotal $c = 3$: $2$.
- Total sub-case (c): $3 + 2 = 5$.
- Hmm, that's $5$ not $6$.
💡 Split by the interior valley value $c \in \{2, 3\}$.
5.OA.B.3 Step 8 - Recount sub-case (a) the same way for consistency.
- Pattern $1 < b > c < d > e$, $\{b, c, d, e\} = \{2, 3, 4, 5\}$, $c$ is the interior valley ($c < b, c < d$), so $c \in \{2, 3\}$.
- $c = 2$: $1 < b > 2 < d > e$, $\{b, d, e\} = \{3, 4, 5\}$.
- Need $b > 2$ (auto), $d > 2$ (auto), $d > e$.
- $d \in \{3, 4, 5\}$ as the peak: $d = 5$: $\{b, e\} = \{3, 4\}$, $e < 5$ auto, $b$ free — $2$ orderings.
- $d = 4$: $\{b, e\} = \{3, 5\}$, $e < 4$ so $e = 3, b = 5$ — $1$.
- $d = 3$: $\{b, e\} = \{4, 5\}$, $e < 3$ impossible — $0$.
- Subtotal $c = 2$: $3$.
- $c = 3$: $1 < b > 3 < d > e$, $\{b, d, e\} = \{2, 4, 5\}$.
- Need $b > 3, d > 3$, so $\{b, d\} = \{4, 5\}$, $e = 2$.
- $b, d$ orderings: $2$.
- Subtotal $c = 3$: $2$.
- Sub-case (a) total: $3 + 2 = 5$.
💡 Mirror of (c) — same count by left-right symmetry.
4.NBT.B.4 Step 9 - Add up-down-up-down (positive start) counts: $5 + 6 + 5 = 16$.
- By Tool #16 symmetry (negate all positions: $a_i \to 6 - a_i$ flips up-down-up-down to down-up-down-up), down-up-down-up also has $16$.
- Total valid: $16 + 16 = 32$.
- Choice (D).
💡 Replace $a_i$ with $6 - a_i$ — every up becomes a down and vice versa, doubling the count.
6.NS.C.7 Reformulate: a triple $a < b < c$ or $a > b > c$ is forbidden. So for every inte 4.OA.B.4 Easier-Problem warmup: try $n = 3$. List all $3! = 6$ permutations of ${1, 2, 3 5.OA.B.3 Easier-Problem step 2: try $n = 4$. Use Tool #16 (symmetry): valid arrangements 5.OA.B.3 Apply the same symmetry to $n = 5$: count arrangements of pattern $a < b > c < d 5.OA.B.3 Sub-case (a): $1$ at position 1, pattern $1 < b > c < d > e$. The four remaining 5.OA.B.3 Sub-case (b): $1$ at position 3, pattern $a < b > 1 < d > e$ with ${a, b, d, e\ 5.OA.B.3 Sub-case (c): $1$ at position 5, pattern $a < b > c < d > 1$. Mirror of sub-case 5.OA.B.3 Recount sub-case (a) the same way for consistency. Pattern $1 < b > c < d > e$, 4.NBT.B.4 Add up-down-up-down (positive start) counts: $5 + 6 + 5 = 16$. By Tool #16 symme Review
Reasonableness: Sanity via the easier-problem pattern: $a_3 = 4, a_4 = 10, a_5 = 32$. The ratio $a_5 / a_4 = 3.2$ and $a_4 / a_3 = 2.5$ — growth is reasonable (well under $n+1$ per step since constraints get stronger as $n$ grows). Another check: the well-known "zigzag" (Euler) numbers $E_n$ satisfy $E_3 = 2, E_4 = 5, E_5 = 16$, and the number of alternating permutations of $\{1, \dots, n\}$ is $2 E_n$. Plugging in: $2 \cdot 2 = 4$, $2 \cdot 5 = 10$, $2 \cdot 16 = 32$ — matches our hand counts at every $n$. Answer $32 = $ (D) confirmed.
Alternative: Tool #2 (Systematic List) brute force — write out all $120$ permutations of $\{1, 2, 3, 4, 5\}$ in lexicographic order and strike through any with three monotone consecutive terms. Tedious but bulletproof, and the resulting count of $32$ is a direct check on the case-split argument above. AoPS-style published enumeration gets the same answer.
CCSS standards used (min grade 6)
6.NS.C.7Understand ordering and absolute value of rational numbers (Reading "three increasing" / "three decreasing" as constraints on the order of adjacent terms.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing all $6$ permutations of $\{1, 2, 3\}$ and identifying the two monotone ones — basic enumeration thinking.)5.OA.B.3Generate two numerical patterns using two given rules and identify relationships (Systematic enumeration of alternating permutations: split by the position of the smallest element, then by interior-valley value.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding subcounts $5 + 6 + 5 = 16$, then doubling to $32$.)
⭐ "No three in a row going up or down" means the arrangement has to zigzag. Pin the smallest number $1$ (it can only be a valley, so positions $1, 3$, or $5$), count carefully — you get $5 + 6 + 5 = 16$ up-down-up-down arrangements. Double for the down-up-down-up mirror pattern: $32$. Answer $\textbf{(D) }32$.
⭐ "No three in a row going up or down" means the arrangement has to zigzag. Pin the smallest number $1$ (it can only be a valley, so positions $1, 3$, or $5$), count carefully — you get $5 + 6 + 5 = 16$ up-down-up-down arrangements. Double for the down-up-down-up mirror pattern: $32$. Answer $\textbf{(D) }32$.
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