Take $\text{AAU}$ together and treat it like $1$ entity. Now arrange $\boxed{\text{AAU }}\text{BCS}$ in ${4!}$ ways.
Then, the $\text{AAU}$ can be arranged in $\dfrac{3!}{2!}$ ways because $A$ has been repeated twice.
So, total arrangements $= \dfrac{4!3!}{2!}$
Option (D) is the correct answer.