How many permutations of a word do not contain consecutive vowels? The word is "ENGINEERING".
The number of ways that the consonants can be ordered is 6! /  3!2!
The number of ways that the vowels can be ordered is 5! / 3!2!
But how would I determine how many ways vowels can be ordered so that they are not next to each other? 
 A: Imagine that you arrange the consonants first. There are six consonants which you can arrange in $6!/(3!2!)$ ways.
Now there are 7 spaces for the 5 vowels to go into but only one vowel can go into each space. So you choose 5 of the 7 available spaces and put a permutation of the vowels into these spaces.
Total number of arrangements with no consectutive vowels $= 6!/(3!2!) \times 5!/(3!2!) \times \binom{7}{5}$.
A: (edit: Answer has been updated to include four missed combinations and the fact that permutations are not considered unique if their spelling matches that of another permutation.)
12345678901  (11 letters in ENGINEERING)
C.C.C.C.C.C (alternating sequence, bounded on both sides by C)
.C.C.C.C.CC  (start single consecutive C pair, one end bounded by C)
.C.C.C.CC.C
.C.C.CC.C.C
.C.CC.C.C.C
.CC.C.C.C.C
CC.C.C.C.C.
C.CC.C.C.C.
C.C.CC.C.C.
C.C.C.CC.C.
C.C.C.C.CC. (count: 10)
.CC.CC.C.C. (start double consecutive C pair)
.CC.C.CC.C.
.CC.C.C.CC.
.C.CC.CC.C.
.C.CC.C.CC.
.C.C.CC.CC. (count: 6)
.CCC.C.C.C. (start triple C string)
.C.CCC.C.C.
.C.C.CCC.C.
.C.C.C.CCC. (count: 4)

In all of these formats, there are 6!/(3!2!) ways to order the C's (i.e. consonants) and 5!/(3!2!) ways to order the .'s (i.e. vowels,) so there should be (21)(6!/3!2!)(5!/3!2!), or 12,600, permutations in which there are no adjacent vowels.
