There are $2$ ways to place the remaining $\text{As}$, $2$ ways to place the remaining $\text{Bs}$, and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{\textbf{(C)}\ 4}$.
Three $\text{A's}$, three $\text{B's}$, and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$