If each of the three operation signs, $+$, $\text{--}$, $\times$, is used exactly ONCE in one of the blanks in the expression
\[5\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}4\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}6\hspace{1 mm}\underline{\hspace{4 mm}}\hspace{1 mm}3\]
then the value of the result could equal
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 19$
There are a reasonable number of ways to place the operation signs, so guess and check to find that $5-4+6 \times 3 = \boxed{\text{(E)}\ 19}$.