2022 AMC 8 Problems/Problem 14

Problem

In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$ s do not appear together?

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$

Solution

All valid arrangements of the letters must be of the form $\textbf{E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E\underline{\hspace{3mm}}E}.$ The problem is equivalent to counting the arrangements of $\textbf{B},\textbf{K},\textbf{P},$ and $\textbf{R}$ into the four blanks, in which there are $4!=\boxed{\textbf{(D) } 24}$ ways.

~MRENTHUSIASM

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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2022 AMC 8 Problems/Problem 14

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