AMC 10 · 2015 · #10
Easy mode Grade 2Put the four letters a, b, c, d in a row, using each letter once. The rule is that two letters next to each other in the alphabet may not sit next to each other in your row. For example, ab and ba are not allowed, because a and b are alphabet neighbors. How many rows follow the rule?
Pick an answer.
AMC 10 2015 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 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).
Givens: The four letters to rearrange are $a$, $b$, $c$, $d$; In the alphabet they sit in order $a,b,c,d$, so the alphabet-adjacent pairs are $ab$, $bc$, $cd$ (in either direction); An arrangement is allowed only if every pair of side-by-side letters is NOT an alphabet-adjacent pair; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 4
Unknowns: The number of valid arrangements of $a,b,c,d$ that obey the rule
Understand
Restated: 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).
Givens: The four letters to rearrange are $a$, $b$, $c$, $d$; In the alphabet they sit in order $a,b,c,d$, so the alphabet-adjacent pairs are $ab$, $bc$, $cd$ (in either direction); An arrangement is allowed only if every pair of side-by-side letters is NOT an alphabet-adjacent pair; Answer choices: (A) 0, (B) 1, (C) 2, (D) 3, (E) 4
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
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.
Execute — Answer: C
2.OA.A.1 Step 1 Turn letters into numbers
- Replace each letter by its alphabet position: $a=1$, $b=2$, $c=3$, $d=4$.
- Two letters are alphabet-adjacent exactly when their numbers differ by $1$.
- So the rule becomes: arrange $1,2,3,4$ in a row so that no two side-by-side numbers differ by $1$.
💡 Numbers make "next to each other in the alphabet" something you can just check with subtraction.
2.OA.A.1 Step 2 Organize by the first letter
- Build the list by deciding the first letter, then seeing what is forced.
- Whatever sits beside a letter cannot be one of its number-neighbors.
- There are four starting letters to try — $a$, $b$, $c$, $d$ — and each branch is short because the rule keeps knocking out choices.
💡 Fixing the first letter splits 24 cases into four small, easy-to-track families.
2.OA.B.2 Step 3 Start with A or D — both collapse
- Start with $a$ ($1$).
- The next letter cannot be $b$ ($2$), so it is $c$ ($3$) or $d$ ($4$).
- If $a,c$: the third letter sits next to $c$ and cannot be $b$ or $d$, but only $b$ and $d$ are left — dead end.
- If $a,d$: the third must avoid $c$ (neighbor of $d$), forcing $b$, then $c$ is last, but $b,c$ differ by $1$ — forbidden.
- So no arrangement starts with $a$.
- Starting with $d$ ($4$) is the mirror image and dies the same way.
💡 An end letter like $a$ or $d$ has too few legal partners, so its branches choke quickly.
2.OA.B.2 Step 4 Start with B or C — exactly one each
- Start with $b$ ($2$): its neighbor cannot be $a$ ($1$) or $c$ ($3$), so the second letter must be $d$ ($4$).
- Next to $d$ we cannot use $c$, so the third is $a$, and $c$ is last: $b,d,a,c$.
- Check the gaps $|2-4|,|4-1|,|1-3| = 2,3,2$ — all safe.
- Starting with $c$ is the mirror image and forces $c,a,d,b$, also safe.
- That is one valid arrangement from $b$ and one from $c$, and the $a$ and $d$ starts gave none, so there are $2$ in total — choice (C).
💡 A middle letter like $b$ or $c$ leaves just enough room for one forced chain to survive.
2.OA.A.1 Replace each letter by its alphabet position: $a=1$, $b=2$, $c=3$, $d=4$. Two le 2.OA.A.1 Build the list by deciding the first letter, then seeing what is forced. Whateve 2.OA.B.2 Start with $a$ ($1$). The next letter cannot be $b$ ($2$), so it is $c$ ($3$) or 2.OA.B.2 Start with $b$ ($2$): its neighbor cannot be $a$ ($1$) or $c$ ($3$), so the seco Review
Reasonableness: 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.
Alternative: Instead of listing, count the bad arrangements with complementary counting: glue an alphabet-adjacent pair together as one block and use inclusion–exclusion over the three pairs $ab,bc,cd$. The blocks give $3\cdot2\cdot3! - (\text{overlaps})$ forbidden arrangements; subtracting from $24$ also lands on $2$. For only four letters, though, the direct organized list is faster and less error-prone.
CCSS standards used (min grade 2)
2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Translating "alphabet-adjacent" into the numeric condition "differ by 1" and applying it while building the organized list by first letter.)2.OA.B.2Fluently add and subtract within 20 using mental strategies (Checking each pair of neighbors by computing the gap between their position numbers and rejecting any gap of exactly 1.)
⭐ 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 the letters into the numbers 1-2-3-4, then arrange them so no two neighbors differ by 1 — only two rows survive.
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