AMC 8 · 2022 · #14

Grade 3 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

permutations-basicsystematic-enumerationcombinations-basic identify-subproblemscasework ↑ Prerequisites: permutations-basic

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

The word 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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

120

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

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.

5 E's + 4 others = 9 letters

2STEP 2

Test the idea on BEE: E _ E has one gap, and B fills it only as EBE — 1 way. The slot method checks out.

BEE → E _ E → EBE (1 way)

3STEP 3

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.

5 E's → 5 - 1 = 4 gaps between consecutive E's

4STEP 4

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.

4 letters ⇔ 4 slots

5STEP 5

Count the orderings: 4 choices, then 3, then 2, then 1, so 4 × 3 × 2 × 1 = 24, which is choice (D).

4 × 3 × 2 × 1 = 24 → (D)

Answer

The total number of distinguishable arrangements of BEEKEEPER (ignoring the constraint) is

\frac{9!}{5!}

= 9 · 8 · 7 · 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.

💡Key takeaway

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!

More like this