In the expression $c\cdot a^b-d$, the values of $a$, $b$, $c$, and $d$ are $0$, $1$, $2$, and $3$, although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$
If $a=0$ or $c=0$, the expression evaluates to $-d<0$. 
If $b=0$, the expression evaluates to $c-d\leq 2$. 
Case $d=0$ remains.
In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$. Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$, where $c\cdot a^b=1\cdot 3^2=\boxed{\mathrm{(D)}\ 9}$.