Let $a$, $b$, $c$, and $d$ be real numbers with $|a-b|=2$, $|b-c|=3$, and $|c-d|=4$. What is the sum of all possible values of $|a-d|$?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 15 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 24$
From $|a-b|=2$ we get that $a=b\pm 2$
Similarly, $b=c\pm3$ and $c=d\pm4$.
Substitution gives $a=d\pm 4\pm 3\pm 2$. This gives $|a-d|=|\pm 4\pm 3\pm 2|$. There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm2$:
$4+3+2=9$, 
$4+3-2=5$, 
$4-3+2=3$, 
$-4+3+2=1$, 
$4-3-2=-1$, 
$-4+3-2=-3$, 
$-4-3+2=-5$, 
$-4-3-2=-9$
Therefore, the only possible values of $|a-d|$ are $9$, $5$, $3$, and $1$. Their sum is $\boxed{\textbf{(D) } 18}$.