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

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 $_{22} P _5 \cdot 21!$


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

Tags:

Combinatorics

Discrete Mathematics

Permutations

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