The total letter arrangements for the word GRACEFUL is 8!.
The question is: What fraction of all arrangements of GRACEFUL have no pair of consecutive vowels?
Now the three vowels in this word are A, E, and U. They can appear consecutively in 3P3 ways so this can be handled in 8! / 3P3 ways. But it is also possible that any two can appear consecutively.
So would the answer be: 8! / (3P3) * 2! * 3?
Best Answer
Since there cannot be consecutive vowels, remove two consonants (we'll add them back in shortly). We can look at the number of ways to arrange $3$ vowels and $3$ consonants, now without the restriction on consecutive vowels. There are $6\choose3$ ways to do this. For any arrangement, add one consonant between the first and second vowel, and another between the second and third.
Now assign vowels and consonants: there are $(3!)(5!)$ ways to do this. The fraction you are looking for is therefore: $$\frac{{6\choose3}(3!)(5!)}{8!}$$