Let's first count the unrestricted arrangement of the letters in CONSTANTINOPLE. There are 14 letters (3 N's, 2 O's, 2 T's, and a single copy of each of the 7 letters C,S,A,I,P,L,E). If all these letters were different (say the N's are colored red white or blue, whereas both the O's and the T's are either red or blue), you would have 14! different arrangements. To each actual arrangement of the (uncolored) letters corresponds exactly 3!2!2! arrangements of the colored letter. Therefore, there are 14!/(3!2!2!) or 3632428800 possible arrangements of the (uncolored) letter in CONSTANTINOPLE. So far so good.
What happens when you don't allow adjacent vowels? You have 5 vowels and 9 consonants. In how many ways can you put P vowels in 2P-1+K positions?
Well, if K is negative you can't do it. If K is zero, you've got only one choice (think about it). In general, you have C(K+P,P) choices, because that's the number of ways you could place the vowels in P+K positions if you did not have any restriction. With the restriction, each vowel is in effect occupying two spaces (itself and the position to its right) except for the last one which is only occupying one space. So, you may as well mentally ignore the P-1 "dead spaces" and see the exact correspondence between the two situations.
Here, you have P=5 and K=5, which means C(10,5) = 252 ways of choosing the positions of the vowels. Once this is done, you have 9!/(3!2!) choices for placing the consonants (since there are 3 identical N's and 2 identical T's) and 5!/2! choices for the vowels (there are two O's). All told, you have 252 ´ 9! ´ 5! / 24 = 457228800 possibilities.
Incidentally, the probability that no two vowels are adjacent in an arrangement of P vowels among N letters (N = 2P-1+K) depends only on N and P, because it's the same as the probability of having no adjacent marks when P squares are randomly marked in a row of N squares. (Putting equiprobable arrangements of vowels in the marked squares and equiprobable arrangements of consonants in the unmarked ones won't change the odds.) This probability is:
C(N-P+1,P) / C(N,P)
With P=5 and N=14, this ratio is simply 252/2002, or 18/143 (about 12.59%) which is an easier way to find the ratio of the above pair of large numbers...