# 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?

Feb 5, 2016

$457228800$

#### 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 $\left\{5\right\}$, $\left\{4 , 1\right\}$, $\left\{3 , 2\right\}$, $\left\{3 , 1 , 1\right\}$, $\left\{2 , 2 , 1\right\}$, $\left\{2 , 1 , 1 , 1\right\}$, $\left\{1 , 1 , 1 , 1 , 1\right\}$. The number of different ways to allocate the parts of the cluster among the $6$ subsequences for each of these clusterings is as follows:

$\left\{5\right\} : 6$

$\left\{4 , 1\right\} : 6 \times 5 = 30$

$\left\{3 , 2\right\} : 6 \times 5 = 30$

$\left\{3 , 1 , 1\right\} : \frac{6 \times 5 \times 4}{2} = 60$

$\left\{2 , 2 , 1\right\} : \frac{6 \times 5 \times 4}{2} = 60$

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

$\left\{1 , 1 , 1 , 1 , 1\right\} : 6$

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 $252 \cdot 60 \cdot 30240 = 457228800$