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




$3! × 3!$




$\dfrac{4! × 3!}{2!}$



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Option(D) is correct

$ABACUS$ is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together.

One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged. 

The number of possible rearrangements is $4!$

The group of 3 vowels contains two $a$'s and one $u$. 

The 3 vowels can rearrange amongst themselves in $\dfrac{3!}{2!}$ ways as the vowel $a$ appears twice.

Hence, the total number of rearrangements in which the vowels appear together are:

$\dfrac{4! × 3!}{2!}$

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