The number of ways to order 26 alphabet letters, no two vowels occurring consecutively

What is the shortest solution to the following problem?

What is the number of ways to order the 26 letters of alphabet so that no two of the vowels a,e,i,o,u occur consecutively?

What I thought is to subtract permutations consisting of 2 vowels, 3 vowels, 4 vowels and 5 vowels occurring consecutively from all permutations 26!. But I could not even find an reasonable answer with that method. It seems to be too long and I always make mistakes.

• I would 1.) Find the number of permutations of consonants, 2.) find the number of permutations of vowels, and 3.) find the number of permissible ways to combine these... specifically each vowel will go between two of the consonants (or at the end) so there are 22 "slots" to be filled by five vowels Feb 25, 2013 at 22:40

Take an ordering of the consonants (there are $21!$ such orderings), for example: $$bcdfghjklmnpqrstvwxyz$$ Enumerate the 22 "slots" between these consonants: $$\_b\_c\_d\_f\_g\_h\_j\_k\_l\_m\_n\_p\_q\_r\_s\_t\_v\_w\_x\_y\_z\_$$

We can choose (respecting order) $_{22} P _5$ lists of five slots to put the vowels $aeiou$ in order.

The answer is therefore $_{22} P _5 \cdot 21!$

• = 161 451 464 537 975 567 155 200 000
– Dan
Jun 24, 2013 at 1:56

Separate it into two cases:

Case 1: No vowel occurs at the end of the list. Then all permutation satifying your condition will be the ones where all the vowels have a consonant to the right of it. Then find the number of ways to pair each vowel with a consonant, then treat the pairs as single objects and multiply by the number of permutations of the remaining objects.

Case 2 is solved similarly except you need to discount one of the vowels and multiply by 5 for your choice of vowel to appear on the end

Hint: You have 26 letters and 5 vowels. Count the number of possible vowel positions by treating the problem as choosing 5 partition points among 21 elements, with order mattering. You can use the classic "dots and lines" approach. Note that you're allowed to put a vowel at the end and beginning, too.

the question should be seen in different way:

number of alphabets = 26

no. of vowels = 5

total no of ways in which 26 letters can be arrange in 26 positions is = 26p26 = 26!

so we have 26! ways in which we can arrange this letter. In this 26! way we got both the ways in which vowels occur together & vowel don't occur together.

so no of ways in which vowels occur together is : no. of positions is = 22 (21 are alphabets & 1 is group of 5 alphabets) so no of ways 22 positions is = 22p22 = 22! so no of ways in 5 vowels is = 5p5 = 5!

so total no of way in which vowel occur = 22! * 5!

so total no of ways in which vowel don't occur together = 26! - 22! * 5!