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.