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?
Best Answer
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 $= 6!/(3!2!) \times 5!/(3!2!) \times \binom{7}{5}$.