If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which

$\begin{tabular}{rr}&\ \texttt{7 a 2}\\ -& \texttt{4 8 b} \\  \hline  &\ \texttt{c 7 3} \end{tabular}$
then $\texttt{a+b+c =}$
$\mathrm{(A) \  }14 \qquad \mathrm{(B) \  }15 \qquad \mathrm{(C) \  }16 \qquad \mathrm{(D) \  }17 \qquad \mathrm{(E) \  }18$
Working from right to left, we see that $2 - b = 3$.  Clearly if $b$ is a single digit integer, this cannot be possible.  Therefore, there must be some borrowing from $a$.  Borrow $1$ from the digit $a$, and you get $12 - b = 3$, giving $b = 9$.
Since $1$ was borrowed from $a$, we have from the tens column $(a-1) - 8 = 7$.  Again for single digit integers this will not work.  Again, borrow $1$ from $7$, giving $10 + (a-1) - 8 = 7$.  Solving for $a$:
$10 + a - 1 - 8 = 7$
$1 + a = 7$
$a = 6$
Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$, giving $c = 2$.
As a check, the problem is $762 - 489 = 273$, which is a true sentence.
The desired quantity is $a + b + c = 6 + 9 + 2 = 17$, and the answer is $\boxed{D}$.