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In how many ways can the letters of the word $\text{ABACUS}$ be rearranged such that the vowels always appear together?

1. $\dfrac{(6+3)!}{2!}$

2. $\dfrac{6!}{2!}$

3. $\dfrac{3!3!}{2!}$

4. $\dfrac{4!3!}{2!}$

5. None of the above

## 2 Answers

Best answer

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.

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

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