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ISI2015-DCG-24

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Group the vowels together and make it a single packet = $OUE$

Rest of the words are: $CMPTR$

Therefore, total words to be grouped including one packet = $(OUE)CMPTR$ = $6!$

But the packet itself can be arranged in $3!$ ways

Hence, answer = $6!3!$

So, option $C$ is the correct answer.
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