Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #20

Grade 6 counting

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

permutations-basicsystematic-enumerationpattern-recognitionsymmetry-argument caseworkeasier-related-problemsymmetry-argument ↑ Prerequisites: permutations-basic

📏 Long solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

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 a₃ = 4, a₄ = 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).

1STEP 1

Restate the forbidden pattern

Reframe the ban: every interior term is a peak or a valley, so difference signs alternate — the arrangement must zigzag.

Pattern: a₁ ≶ a₂ ＞ rless a₃ ≶ a₄ ＞ rless a₅

2STEP 2

Warm up with n = 3

Warm up with n = 3: of the 3! = 6 permutations only the two monotone ones (123, 321) are banned, leaving 4 arrangements.

Count₃ = 4

3STEP 3

Try n = 4

For n = 4, symmetry pairs up-down with down-up; listing the up-down-up permutations gives 5, so doubling yields 10.

Count₄ = 2 · 5 = 10

4STEP 4

Split n = 5 by where 1 sits

Now n = 5, pattern a ＜ b ＞ c ＜ d ＞ e: since 1 is below both neighbors it must be a valley — positions 1, 3, or 5.

1 ∈ {pos 1, 3, 5}

5STEP 5

Count case a

1 at position 1 (1 ＜ b ＞ c ＜ d ＞ e): a first count over the interior valley c gives 6 arrangements, revisited in the final step.

Sub-case (a): 6 arrangements with 1 at position 1

6STEP 6

Count case b

1 at position 3 (a ＜ b ＞ 1 ＜ d ＞ e): split {2, 3, 4, 5} into an ascending pair and a descending pair — C(4, 2) = 6 arrangements.

Sub-case (b): 6 arrangements with 1 at position 3

7STEP 7

Count case c

1 at position 5 (a ＜ b ＞ c ＜ d ＞ 1): split by the interior valley c ∈ {2, 3} — c = 2 gives 3 and c = 3 gives 2, for 5 arrangements.

Sub-case (c): 5 arrangements with 1 at position 5

8STEP 8

Recount case a, then add

Case a recounts to 5, not 6, so the up-down total is 5 + 6 + 5 = 16; the mirror gives 16 more — 32 → (D).

Sub-case (a) corrected: 5 arrangements Total = 2 · 16 = 32 → (D)

Answer

Sanity via the easier-problem pattern: a₃ = 4, a₄ = 10, a₅ = 32. The ratio a₅ / a₄ = 3.2 and a₄ / a₃ = 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 E₃ = 2, E₄ = 5, E₅ = 16, and the number of alternating permutations of {1, …, n} is 2 E_n. Plugging in: 2 · 2 = 4, 2 · 5 = 10, 2 · 16 = 32 — matches our hand counts at every n. Answer 32 = (D) confirmed.

💡Key takeaway

"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 (D) 32.

Restate the forbidden pattern
Warm up with n = 3
Try n = 4
Split n = 5 by where 1 sits
Count case a
Count case b
Count case c
Recount case a to check

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