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Best answer
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28 votes

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.

edited by
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Total no of ways = C(4 ,3) ⨉ 3! /2!  ⨉ 3! = 4 ⨉ 3! ⨉ 3!/2! = 4! ⨉3! / 2!

The correct answer is (D) 4! ⨉3! / 2!

Answer:

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