There are $5$ $3$s in total now with other $5$ digits remaining.  There are $\_5P\_5$ ways for us to arrange the $5$ letters\textquotesingle{} positions. Then, we can put the $5$ $3$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 do not appear together? 

- $$1800$$ 
- $$1200$$ 
- $$1000$$ 
- $$720$$ 
- $$120$$ 