# The letters of the word CONSTANTINOPLE are written on 14

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?
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Step 1
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:
$\left\{5\right\}:6$
$\left\{4,1\right\}:6×5=30$
$\left\{3,2\right\}:6×5=30$
$\left\{3,1,1\right\}:\frac{6×5×4}{2}=60$
$\left\{2,2,1\right\}:\frac{6×5×4}{2}=60$
$\left\{2,1,1,1\right\}:\frac{6×5×4×3}{3!}=60$
$\left\{1,1,1,1,1\right\}:6$
Step 2
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 $\frac{5!}{2!}=60$ ways since there are 2O's.
The 9 consonants can be ordered in $\frac{9!}{3!2!}=30240$ ways since there are 3 N's and 2T's
So the total possible number of arrangements satisfying the conditions is $252\cdot 60\cdot 30240=457228800$.