Question

The letters of the word CONSTANTINOPLE are written on 14 cards, one of each card. The cards are shuffled and then arranged in a straight line. How many arrangements are there where no two vowels are next to each other?

Answer 1

`457228800`

Explanation:

CONSTANTINOPLE

First of all just consider the pattern of vowels and consonants.

We are given `5` vowels, which will split the sequence of `14` letters into `6` subsequences, the first before the first vowel, the second between the first and second vowels, etc.

The first and last of these `6` sequences of consonants may be empty, but the middle `4` must have at least one consonant in order to satisfy the condition that no two vowels are adjacent.

That leaves us with `5` consonants to divide among the `6` sequences. The possible clusterings are `{5}`, `{4,1}`, `{3,2}`, `{3,1,1}`, `{2,2,1}`, `{2,1,1,1}`, `{1,1,1,1,1}`. The number of different ways to allocate the parts of the cluster among the `6` subsequences for each of these clusterings is as follows:

`{5}: 6`

`{4,1}: 6xx5 = 30`

`{3,2}: 6xx5 = 30`

`{3, 1, 1}: (6xx5xx4)/2 = 60`

`{2, 2, 1}: (6xx5xx4)/2 = 60`

`{2, 1, 1, 1}: (6xx5xx4xx3)/(3!) = 60`

`{1,1,1,1,1}: 6`

That is a total of `252` ways to divide `5` consonants among `6` subsequences.

Next look at the subsequences of vowels and consonants in the arrangements:

The `5` vowels can be ordered in `(5!)/(2!) = 60` ways since there are `2` O's.

The `9` consonants can be ordered in `(9!)/(3!2!) = 30240` ways since there are `3` N's and `2` T's

So the total possible number of arrangements satisfying the conditions is `252*60*30240 = 457228800`