There are $5$ $3$s in total.  There are $\_5P\_5$ ways for us to arrange the other $5$ letters\textquotesingle{} positions. Then, we can put the five $3$s in the $6$ intervals.  So the answer is $\_5P\_5 \times \_6C\_5=720$. 
In how many ways can the digits in $3433256337$ be rearranged so that two or more $3$s are not adjacent to each other? 

A) $$1800$$ 
B) $$1200$$ 
C) $$1000$$ 
D) $$720$$ 
E) $$120$$ 