There are $3$ $E$s in total now with other $8$ letters remaining. But pay attention to $B$ and $P$: there are $3$ $B$s and $3$ $P$ here.  There are $\_8P\_5 \div \_3P\_3$ ways for us to arrange the $8$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $9$ intervals.  So the answer is $\_8P\_5 \div \_3P\_3 \times \_9C\_3=94080$. 
In how many ways can the letters in $BEEKBBPERPP$ be rearranged so that two or more $E$s do not appear together? 
Options:
a) $$49200$$ 
b) $$94080$$ 
c) $$564480$$ 
d) $$1800$$ 
e) $$98400$$ 