In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $\textbf{E}$s do not appear together?
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 24 \qquad \textbf{(E) } 120$
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