How many arrangements of GRACEFUL... How many arrangements of GRACEFUL have no pair of consecutive vowels? 
Soln:
Breaking up the question into pieces: 
First I found the total amount of arrangememts possible with GRACEFUL:  $$8!$$
Next I went about finding all the possible arrangements with consecutive vowels:  Letting V'= double vowel and V = single vowel I am left with the string:
                   "VV'GRCFL"

This could be arramged in $$(7!)(3)(2)$$ types of ways. 
Therefore to find arrangements with no consecutive vowels: $$8! - (7!)(3)(2)$$ Where 3 was for the ways the single vowel could be chosen and 2 was for the possible arrangements of the double vowel
Initially I had split the consecutive vowels into a set of 3 and the sets of doubles,  but I realized I could count all arramgements together.  I feel there may be an easier way using the method I applied which was with some assistance from the text.  
Did I approach correctly?  Ways to improve? 
Edit: The solution by N.F Taussig is a very well written explanation that really provides a visual of what I did wrong and would serve as a guide for others
 A: Your answer is incorrect because you excluded those arrangements in which all three vowels are consecutive twice.
Method 1: Placing vowels in the spaces between consonants or at the ends.
We can arrange the five distinct consonants in $5!$ ways.  For a particular arrangement of the consonants, we have six spaces in which to place to the vowels (four between successive consonants and the two ends of the row).  Therefore, we can place the A in six ways, the E in five ways, and the U in four ways.  Hence, the number of permutations in which no pair of vowels is consecutive is $$5! \cdot 6 \cdot 5 \cdot 4$$
Method 2:  We correct your attempt by using the Inclusion-Exclusion Principle.
As you found, there are $8!$ ways to arrange the letters of the word GRACEFUL.  From these, we must exclude those arrangements in which at least two vowels are consecutive.  
Suppose that two vowels are consecutive.  Then we have seven objects to arrange, the five consonants, the double vowel, and the single vowel.  There are $\binom{3}{2}$ ways of choosing two of the three vowels to be in the double vowel and $2!$ ways of arranging the chosen vowels within the double vowel.  Since there are $7!$ ways of arranging seven distinct objects, there are $$\binom{3}{2} \cdot 2! \cdot 7! = 3 \cdot 2 \cdot 7! = 6 \cdot 7!$$
arrangements with at least two distinct vowels, which you correctly calculated.
However, if we exclude arrangements with at least two consecutive vowels, we have excluded arrangements with three consecutive vowels twice, once when we exclude the first two of the three consecutive vowels and once when we exclude the last two of the three consecutive vowels.  Since we only want to exclude arrangements with three consecutive vowels once, we must add the number of arrangements with three consecutive vowels.
Suppose that three vowels are consecutive.  Then we have six distinct objects to arrange, the five consonants and the block of three vowels.  The block of three distinct vowels can be arranged internally in $3!$ ways.  The six objects can be arranged in $6!$ ways.  Hence, the number of arrangements in which three vowels are consecutive is $$3!6!$$
By the Inclusion-Exclusion Principle, the number of arrangements of the letters of the word GRACEFUL in which no two letters are consecutive is $$8! - 6 \cdot 7! + 3!6!$$
A: Total ways are $8!$ now consider A,E as a block so total ways become $(1+6)! \cdot 2!=10080$ same logic goes for  A,U and E,U so total ways are $8!-[(3 \cdot 7!) \cdot 2!)]=40320-30240=10080$ so total ways are $10080$ . now let us three different vowels as a block so it becomes $(1+5)!.3!=4320$ so total ways become $10800+4320=14400$
