TITLE: How many permutations of a word do not contain consecutive vowels?
QUESTION [5 upvotes]: 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?

REPLY [8 votes]: 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 $= 6!/(3!2!) \times 5!/(3!2!) \times \binom{7}{5}$.