Suppose there are $$x$$ red marbles, $$y$$ green marbles, and $$z$$ blue marbles. Thus, we can get  $$\begin{cases} x+y=4① \textbackslash\textbackslash{} y+z=6②\textbackslash\x+z=8③\end{cases}$$,  ①$$+$$②$$+$$③: $$2x+2y+2z=18$$, $$x+y+z=9$$. 
In a jar of red, green, and blue marbles, all but $$6$$ are red marbles, all but $$8$$ are green, and all but $$4$$ are blue. How many marbles are in the jar? ($$2012$$ AMC $$8$$ problem, Question \#$$19$$) 
A. $$6$$ 
B. $$8$$ 
C. $$9$$ 
D. $$10$$ 
E. $$12$$ 