Competition · AMC preparation · step 4 of 4
AMC 8 · 2022 · #14
Grade 3 countingPick an answer.
AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 BEE first to confirm the slot-counting logic.
Count the letters
BEEKEEPER has 5 E's plus 4 distinct letters B, K, P, R — nine in all. Only the E's are constrained, so handle them first.
Counting and combining two groups of letters is a Grade 2 addition word-problem skill.
2.OA.A.1Identify SubproblemsTry a shorter word first
Test the idea on BEE: E _ E has one gap, and B fills it only as EBE — 1 way. The slot method checks out.
Trying a smaller version of the same situation is a Grade 3 word-problem habit and shows the slot method works.
3.OA.A.3Solve An Easier Related ProblemCount the gaps
Spread the E's out: E _ E _ E _ E _ E. Five E's leave exactly 4 gaps, and one letter per gap keeps every pair of E's apart.
Counting the spaces between 5 objects is a Grade 1 "one less" subtraction observation.
1.OA.A.1Draw A DiagramMatch letters to gaps
The four letters B, K, P, R match the 4 slots one-to-one, so now just count the orderings of these four distinct letters.
Recognizing that the original counting problem reduces to a smaller, equivalent counting problem is the Tool #7 "subproblem" move.
3.OA.A.3Identify SubproblemsCount the orderings
Count the orderings: 4 choices, then 3, then 2, then 1, so 4 × 3 × 2 × 1 = 24, which is choice (D).
Multiplying 4 × 3 × 2 × 1 is Grade 3 "fluently multiply within 100".
The number of rearrangements of BEEKEEPER in which no two E's stand next to each other equals 4 × 3 × 2 × 1 = 24.
▸ Why?
Every such rearrangement is exactly the five E's set in a row with one of the four other letters dropped into each gap between neighboring E's, so each valid rearrangement pairs with one ordering of B, K, P, R and each ordering rebuilds one valid rearrangement — making the two collections equal in size.
▸ Why?
The pairing lines up perfectly because five E's in a row leave exactly four gaps between neighbors — one gap for every E except the first — and there are exactly four non-E letters to fill them, and a filled gap between each pair of E's is just what stops any two E's from touching.
▸ Why?
Ordering the four distinct letters, the first spot has 4 choices, the second has 3 left, the third 2, and the last 1, and the number of full orderings is the product 4 × 3 × 2 × 1.
▸ Why?
Filling the spots is four choices made one after another — 4 options, then 3, then 2, then 1 — and how many letters remain for a later spot does not depend on which letters landed in the earlier spots, so these independent successive choices multiply.
This AMC 8 problem only needs Grade 3 multiplication (4 × 3 × 2 × 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!
- Count the letters
- Try a shorter word first
- Count the gaps
- Match letters to gaps
- Count the orderings
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