There are $3$ $E$s in total now with other $6$ letters remaining. But pay attention to $B$: there are $3$ $B$s here.  There are $\_6P\_3$ ways for us to arrange the $6$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $7$ intervals.  So the answer is $\_6P\_3 \times \_7C\_3=4200$. 
In how many ways can the letters in $BEEKBBPER$ be rearranged so that two or more $E$s do not appear together? 

  - $$4200$$ 
  - $$900$$ 
  - $$800$$ 
  - $$720$$ 
  - $$700$$ 