Random arrangement of English Alphabet; Finding probability that no two vowels are next to each other. So, in a random arrangement of the alphabet, the sample space will be $26^{26}$.
So, $P(\text{Sample Space}) = 26^{26}$
No two vowels next to each other might mean, consonants on 13 spaces and vowels on 13 slots.
Here's what I did:
Consider the event $A$.
$N(A) = 13^{21}$ ($13$ Slots for $21$ consonants, randomly). $13^5$ ($13$ slots for random $5$ vowels).
I am not getting the permutation part quite well
So, how do I proceed?
 A: Addendum added to respond to the comment of Jean Marie.

I am assuming that there are exactly $(5)$ vowels in the alphabet.  For $k \in \{0,1,2,3, \cdots, 13\}$, let $f(k)$ denote the number of ways that you can combine $(k)$ vowels and $26-k$ consonants into a $(26)$ character string where:

*

*Replacement is permitted  (i.e. you can re-use a letter).


*No two vowels can be together.
Then, the desired enumeration will be
$$\sum_{k=0}^{13} f(k), \tag1 $$
so, the problem has been reduced to providing a closed form expression for $f(k)$.
For a specific value of $(k)$, you will have $5$ choices for each of the $k$ vowel-positions and $21$ choices for each of the $(26-k)$ consonant positions.
Therefore, to enumerate $f(k)$, the first thing that is necessary is to create the factors
$$(21)^{(26 - k)} \times (5)^k. \tag2 $$
The factors in (2) above will be the first two factors used in the enumeration of $f(k)$.
For the sake of illustration, assume that $k = 10,$ and that you are (arbitrarily) going to form a 26 character word with $(16)$ Z's, and $(10)$ A's.  How many such 26 character words are possible, where none of the A's are next to each other.
To answer this question, you need a working knowledge of Stars and Bars Theory.
For Stars and Bars theory, see
this article and
this article.
Consider the number of non-negative integer solutions to
$$x_1 + x_2 + \cdots + x_{11} = 16. \tag3 $$
This enumeration, which per Stars and Bars Theory is
$~\displaystyle \binom{16 + 10}{10}~$ is close but slightly different from the number of satisfying ways of combining the $10$ A's and $(16)$ Z's.
One good thing about the equation in (3) is that it respects the $(16)$ choices of consonants, which are to be placed around the $(10)$ vowels.  The other good thing is that with $(10)$ vowel positions to be used, the variables $x_1,\cdots,x_{11}$ nicely represent the gaps before the first vowel, after the last vowel, and between the vowels.
The only bad thing is that while it is okay for $x_1$ and $x_{11}$ to equal $(0)$, because these variables represent the gaps before the first vowel and after the last vowel, it is not okay for any of the other variables to equal $(0)$.
For example, if $x_2 = 0$, this analogizes to there being no gap between the first and second vowel.
The fix here is to change the variables:
For $k \in 2,3,\cdots,10$, let $y_k = x_k - 1.$
Then, you are interested in the number of non-negative integer solutions to
$x_1 + $ 
$y_2 + y_3 + \cdots + y_{10}$ 
$+ x_{11} = (16 - 9) = 7.$
So, for the specific use of $16$ Z's and $10$ A's, the enumeration is
$$\binom{7 + 10}{10}.$$
In general, for $k \in \{0,1,2,\cdots,13\}$, you will want the number of non-negative integer solutions to
$$x_1 + x_2 + \cdots + x_{k+1} = (26 - k)$$
subject to the constraint that none of $x_2,x_3, \cdots, x_k$ are permitted to equal $(0)$.
The enumeration here is
$$\binom{[26 - k] + [k+1 - 1] - [k-1]}{k+1 - 1} = \binom{27-k}{k}. \tag4 $$

Putting it all together, the overall enumeration is
$$\sum_{k=0}^{13} ~\left[ ~(21)^{(26 - k)} \times (5)^k \times \binom{27-k}{k} ~\right]. \tag5 $$
So, the expression in (5) above represents the enumeration of the number of satisfying (equally likely) $(26)$ character strings.
To convert the enumeration into a Probability, simply multiply it by
$$\frac{1}{26^{(26)}}.$$

Addendum 
Responding to the comment of Jean Marie:

As I remarked above, there are 6 vowels in the alphabet A,E,I,O,U and Y.

Except that $Y$ is sometimes uses as a consonant (e.g. the word "yardstick") and sometimes used as a vowel (e.g. the word "my").
To make sense out of the problem, a decision must be made, one way or another.  The OP (i.e. Original Poster) left a comment that he thought that there were $(5)$ or $(6)$ vowels.
That doesn't cut it.  You can't ask the responder to try and hit a moving target.  Normally, I try to anticipate the intent of the OP.
Here, there is certainly enough evidence in support of your viewpoint to justify including the alternative computation.
The only relevance to altering the number of vowels from $(5)$ to $(6)$ is that the factors of
$$(21)^{(26-k)} \times (5)^k$$
change to
$$(20)^{(26-k)} \times (6)^k.$$
A: (Assuming repetition is allowed)
Here's another way to count it. The number of allowed arrangements is given by sum of first row of
$$
\begin{bmatrix}
21 & 5 \\
21 & 0
\end{bmatrix} ^ {26}
$$
which is $2727004680535608837987658411692498651$. This follows when you consider it as a state machine, where states are $C = $ "previous was a consonant" and $V = $"previous was a vowel". Then the number of ways you can transition are indicated in the matrix, i.e you can append a consonant in $21$ ways and a vowel in $5$ ways. And if previous was a vowel, you can't append a vowel (hence the $0$ in the matrix). You start from the state $C$ because you can assume "previous" was a consonant when there's nothing (it makes no restrictions). Then take $26$ steps, i.e. append $26$ letters.
