A bag contains four pieces of paper, each labeled with one of the digits $1$, $2$, $3$ or $4$, with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3$?
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{3}{4}$
The number of ways to form a 3-digit number is $4 \cdot 3 \cdot 2 = 24$. The combination of digits that give us multiples of 3 are (1,2,3) and (2,3,4), as the integers in the subsets have a sum which is divisible by 3. The number of 3-digit numbers that contain these numbers is $3! + 3! = 12$. 
Therefore, the probability is $\frac{12}{24} = \boxed{\frac{1}{2}}$.
~abc2142