Combinations word "GRACEFUL" question 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`?

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: $$\frac{{6\choose3}(3!)(5!)}{8!}$$