AMC 8 · 2022 · #14
Easy mode Grade 3Problem
Picture the word BEEKEEPER written out. It has 9 letters in total. Five of them are the letter E. The other four are B, K, P, and R.
We want to rearrange all 9 letters into a new order. There is one rule. No two Es are allowed to sit next to each other.
How many different orderings follow this rule?
Pick an answer.
AMC 8 2022 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: The word $\textbf{BEEKEEPER}$ has $9$ letters: five $E$'s and four other distinct letters $B$, $K$, $P$, $R$. We must count rearrangements (orderings of all $9$ letters) in which no two $E$'s sit next to each other.
Givens: Letter multiset: $E \times 5$, plus $B, K, P, R$ once each ($9$ letters total); Constraint: no two $E$'s may be adjacent in the final arrangement; Answer choices: (A) $1$, (B) $4$, (C) $12$, (D) $24$, (E) $120$
Unknowns: The number of valid rearrangements of $\textbf{BEEKEEPER}$ that keep every pair of $E$'s separated
Understand
Restated: The word $\textbf{BEEKEEPER}$ has $9$ letters: five $E$'s and four other distinct letters $B$, $K$, $P$, $R$. We must count rearrangements (orderings of all $9$ letters) in which no two $E$'s sit next to each other.
Givens: Letter multiset: $E \times 5$, plus $B, K, P, R$ once each ($9$ letters total); Constraint: no two $E$'s may be adjacent in the final arrangement; Answer choices: (A) $1$, (B) $4$, (C) $12$, (D) $24$, (E) $120$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
The constraint "no two $E$'s adjacent" is hard to attack head-on, but it becomes obvious once we DRAW the $5$ $E$'s spread out in a row with gaps between them: $E\,\_\,E\,\_\,E\,\_\,E\,\_\,E$. Tool #1 (Draw a Diagram) makes the structure visible — there are exactly $4$ gaps and exactly $4$ non-$E$ letters, so the problem instantly turns into "arrange $B, K, P, R$ in $4$ slots". Tool #7 (Identify Subproblems) names that decomposition: (a) lock the $E$ skeleton, (b) permute the $4$ distinct letters. Tool #9 (Easier Problem) is the sanity check — try a shorter word like $\textbf{BEE}$ first to confirm the slot-counting logic.
Execute — Answer: D
2.OA.A.1 Step 1 - Inventory the letters of $\textbf{BEEKEEPER}$.
- There are $5$ $E$'s and $4$ other distinct letters: $B$, $K$, $P$, $R$ — nine letters in total.
- The constraint only affects the $E$'s, so it makes sense to handle them first.
💡 Counting and combining two groups of letters is a Grade 2 addition word-problem skill.
3.OA.A.3 Step 2 - Try the SAME idea on a tiny word first.
- For $\textbf{BEE}$ ($2$ $E$'s, $1$ other letter $B$), draw $E \,\_\, E$ — exactly $1$ gap.
- Only $1$ way to drop $B$ in: $EBE$.
- Indeed the only valid arrangement of $\textbf{BEE}$ with no two $E$'s adjacent is $EBE$.
- The mini-experiment confirms the strategy: place the $E$'s spread out, then fill the gaps.
💡 Trying a smaller version of the same situation is a Grade 3 word-problem habit and shows the slot method works.
1.OA.A.1 Step 3 - Now draw the diagram for the full problem.
- Place the $5$ $E$'s in a row with one blank slot between each pair: $E \,\_\, E \,\_\, E \,\_\, E \,\_\, E$.
- Between the $5$ $E$'s there are exactly $5 - 1 = 4$ gaps.
- As long as each gap gets a non-$E$ letter, NO two $E$'s can sit next to each other — the gap letter separates them automatically.
💡 Counting the spaces between $5$ objects is a Grade 1 "one less" subtraction observation.
3.OA.A.3 Step 4 - Match the $4$ non-$E$ letters to the $4$ gaps.
- We have exactly $4$ different letters ($B, K, P, R$) and exactly $4$ open slots.
- Every valid arrangement of $\textbf{BEEKEEPER}$ that meets the constraint corresponds to one way of dropping $B, K, P, R$ into those $4$ slots, and vice versa.
- So the question reduces to: in how many orders can we line up the $4$ distinct letters $B, K, P, R$?
💡 Recognizing that the original counting problem reduces to a smaller, equivalent counting problem is the Tool #7 "subproblem" move.
3.OA.C.7 Step 5 - Count the orderings of $B, K, P, R$.
- The first slot can be filled $4$ ways, the second $3$ ways (one letter used), the third $2$ ways, and the last $1$ way.
- Multiply: $4 \times 3 \times 2 \times 1 = 24$.
- This matches answer choice $\textbf{(D)}$.
💡 Multiplying $4 \times 3 \times 2 \times 1$ is Grade 3 "fluently multiply within $100$".
2.OA.A.1 Inventory the letters of $\textbf{BEEKEEPER}$. There are $5$ $E$'s and $4$ other 3.OA.A.3 Try the SAME idea on a tiny word first. For $\textbf{BEE}$ ($2$ $E$'s, $1$ other 1.OA.A.1 Now draw the diagram for the full problem. Place the $5$ $E$'s in a row with one 3.OA.A.3 Match the $4$ non-$E$ letters to the $4$ gaps. We have exactly $4$ different let 3.OA.C.7 Count the orderings of $B, K, P, R$. The first slot can be filled $4$ ways, the Review
Reasonableness: The total number of distinguishable arrangements of $\textbf{BEEKEEPER}$ (ignoring the constraint) is $\tfrac{9!}{5!} = 9 \cdot 8 \cdot 7 \cdot 6 = 3024$, so an answer like $24$ is a small fraction of all arrangements — which feels right because the "no two $E$'s touching" rule is quite restrictive (with $5$ $E$'s out of $9$ slots, almost all arrangements have some $E$'s touching). Also, the answer is exactly $4! = 24$, which matches our intuition that the $E$ skeleton is forced and only the four distinct letters can shuffle. Choice (E) $120 = 5!$ would correspond to a five-slot permutation, but we have $4$ slots, not $5$. Choice (D) $24$ is correct.
Alternative: Tool #16 (Change Focus / Count the Complement) works too but is messier here: count ALL distinguishable arrangements of $\textbf{BEEKEEPER}$ ($\tfrac{9!}{5!} = 3024$) and subtract those with at least one $EE$ block — but the complement involves inclusion-exclusion across $EE$, $EEE$, $EEEE$, $EEEEE$ patterns. Direct construction with Tool #1 (Diagram) avoids that headache entirely. Alternatively, Tool #2 (Systematic List) on the toy version $\textbf{BEE} \to EBE$ already shows the slot method gives the unique answer; scaling up gives $4! = 24$ with the same logic.
CCSS standards used (min grade 3)
1.OA.A.1Solve addition and subtraction word problems within 20 (Computing $5 - 1 = 4$ gaps between the $5$ $E$'s — the structural observation that drives the whole solution.)2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Inventorying the letters: $5$ $E$'s plus $4$ other distinct letters add to $9$ total.)3.OA.A.3Solve multiplication and division word problems within 100 (Recognizing the problem reduces to "$4$ distinct letters into $4$ slots" and validating it on the tiny $\textbf{BEE}$ case.)3.OA.C.7Fluently multiply and divide within 100 (Calculating $4 \times 3 \times 2 \times 1 = 24$ to count the orderings of $B, K, P, R$.)
⭐ This AMC 8 problem only needs Grade 3 multiplication ($4 \times 3 \times 2 \times 1 = 24$) you already know — once you draw the picture $E\_E\_E\_E\_E$, the rest is just arranging $4$ letters in $4$ blanks!
⭐ This AMC 8 problem only needs Grade 3 multiplication ($4 \times 3 \times 2 \times 1 = 24$) you already know — once you draw the picture $E\_E\_E\_E\_E$, the rest is just arranging $4$ letters in $4$ blanks!
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