Consider the word UNBIASED. How many words can be formed with the letters of the word in which no two vowels are together?
1 Answer
2880 ways.
Explanation:
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! 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.