In how many ways can we arrange the $26$ English letters in a row so that no two vowels are adjacent to each other, and each block of consonant(s) (between $2$ vowels) is/are in alphabetical order?
Example of a valid arrangement:
$\underset{*}{\underbrace{RX}}E\underset{*}{\underbrace{BGMQ}}O\underset{*}{\underbrace{CDJY}}I\underset{*}{\underbrace{FKNZ}}U\underset{*}{\underbrace{HTV}}A\underset{*}{\underbrace{LPSW}}$
The $*$ means that the consonants are in alphabetical order, so it is valid.
Example of invalid arrangement:
$I\underset{\text{invalid}}{\underbrace{BRMQGW}}O\underset{*}{\underbrace{CDJY}}A\underset{*}{\underbrace{FKNZ}}U\underset{*}{\underbrace{HTV}}E\underset{*}{\underbrace{LPSX}}$
$R$ can not be before $M$, and $M$ can not be before $G$.
The result of my trial:
$\frac{1}{2}(\frac{26!}{21!})=26 \times 25 \times 24 \times 23 \times 22 \times \frac{1}{2}=3,946,800$ different arrangments. I am not sure about it. Please tell me if I am wrong.
Unfortunately, I have no knowledge about inclusion-exclusion principle.
Any help would be appreciated. THANKS!
I assume that we consider the list of vowels to be only $A,E,I,O,U$ and we do not consider $Y$ to be a vowel.
We use here Stirling Numbers of the Second Kind. We use $\left\{\begin{matrix}n\\k\end{matrix}\right\}$ to denote the number of partitions of an $n$ element set into $k$ non-empty unlabeled subsets.
We recognize that a vowel may appear at the very front of the arrangement and/or at the very end, and so we may have either $4,5$ or $6$ blocks of consonants. In each block of consonants, we will require that they appear in alphabetical order.
Once having partitioned our consonants into $k$ non-empty subsets, we may then decide in what order they appear and which (if any) of the blocks to the far left of the furthest left vowel or to the far right of the furthest right vowel are unused.
We then decide which vowel appears in which position separating the blocks.
We have then a final total of:
$$\left(\left\{\begin{matrix}21\\4\end{matrix}\right\}4! + 2\cdot \left\{\begin{matrix}21\\5\end{matrix}\right\}5! + \left\{\begin{matrix}21\\6\end{matrix}\right\}6!\right)5!$$
Evaluated this becomes approximately $2.4\times 10^{18}$