AMC 10 · 2015 · #10

Grade 2 arithmetic
Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count
systematic-enumerationpermutations-basic casework ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
Count how many orderings of the four letters a, b, c, d have the property that no two letters sitting next to each other are also next to each other in the alphabet (so neighbors like ab, ba, bc, cb, cd, dc are forbidden).

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

There are only 4!=24 arrangements, a small finite world, so Tool #2 (Make a Systematic List) — organized by which letter comes first — guarantees we miss none and double-count none. To make the alphabet-adjacency test mechanical, Tool #4 (Introduce a Variable) relabels a,b,c,d as positions 1,2,3,4, turning "alphabet-adjacent" into the crisp numeric test "differ by exactly 1." Then Tool #3 (Eliminate Possibilities) prunes whole branches the moment a forbidden pair is forced, so most of the 24 cases die instantly instead of being written out.

1STEP 1

Turn letters into numbers

Map letters to numbers a=1,b=2,c=3,d=4; alphabet-adjacent means the numbers differ by 1.

a=1, b=2, c=3, d=4; forbidden neighbor ⇔ |difference|=1
2STEP 2

Organize by the first letter

Try each starting letter and see what the ban forces for the rest of the row.

first letter ∈ {a,b,c,d}, then fill the rest under the "differ by 1" ban
3STEP 3

Start with A or D — both collapse

Starting with a or d always dead-ends: every branch is forced into a forbidden neighbor pair.

a c ?→{b,d} both touch c; a d b c ends in bc — both fail
4STEP 4

Start with B or C — exactly one each

Starting with b or c forces exactly one valid row each, bdac and cadb, for a total of 2 — choice (C).

b d a c and c a d b work → total =2 → (C)
Answer
2
The two survivors bdac and cadb are reverses of each other, which fits the symmetry of the problem: flipping a valid row left-to-right keeps every neighbor pair the same, so valid arrangements come in mirror pairs. That makes an even count expected, and 2 is the smallest nonzero even count — ruling out the odd options (B) 1 and (D) 3. A quick recheck of bdac gives neighbor gaps 2,3,2 and cadb gives 2,3,2, both clean, while every a- or d-start was forced into an ab/bc/cd pair. So (A) 0 is too few and (E) 4 is too many.
💡Key takeaway

Turn the letters into the numbers 1-2-3-4, then arrange them so no two neighbors differ by 1 — only two rows survive.

  • Turn letters into numbers
  • Organize by the first letter
  • Start with A or D — both collapse
  • Start with B or C — exactly one each

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