Consider the word UNBIASED. How many words can be formed with the letters of the word in which no two vowels are together?

First we notice that all 8 letters are different. There are no duplcates.

Break it down into two parts:
1. How many ways can 4 vowels be arranged in an 8 letter word so that no two vowels are touching?
2. Within each of those arrangements, how many ways can I position 4 unique vowels and 4 unique consonants?

The one big question is now two smaller, easier questions.

Part 1:

Using X for a consonant and O for a vowel, we can arrange the 4 O's so they're not touching in the following ways:

#"O X O X O X O X"#
#"O X O X O X X O"#
#"O X O X X O X O"#
#"O X X O X O X O"#
#"X O X O X O X O"#

So there are 5 total ways to arrange any 4 vowels within an 8-letter word so that no two vowels are consecutive.

Part 2:

For each of these 5 ways, we can permute 4 unique vowels in #4!# ways, and the 4 unique consonants in #4!# ways as well. That gives

#4! xx 4!" "=" "24xx24" "=" "576#

576 acceptable ways to permute all 8 letters within any of the 5 arrangements above.

Finally:

Since we have a choice of 5 arrangements, and 576 letter permutations within each arrangement, the final number of words we can form that meet our conditions is the product of these two numbers.

#5 xx 576 = 2880#

So there are 2880 total ways to arrange the letters of UNBIASED so that no two vowels are together.