There are $3$ $C$s in total now with other $4$ letters remaining.  There are $\_4P\_4$ ways for us to arrange the remaining $4$ letters\textquotesingle{} positions.  So the answer is $\_4P\_4\times \_5C\_3 =240$. 
In how many ways can the letters in $CPCCKBY$ be rearranged so that two or more $C$s are not adjacent to each other? 

A. $$240$$ 
B. $$180$$ 
C. $$64$$ 
D. $$96$$ 
E. $$140$$ 