Exercise
Find the number of arrangements of the $26$ letters of the English alphabet that have both
- Y occurring before X
- all $5$ vowels (A, E, I, O, U) occurring in alphabetical order
Warm-up
First I count how many arrangements meet the first condition. Probably not necessary, but I am wondering if I am using the right reasoning. Such an arrangement would look like
$$\begin{matrix}[\alpha \text{ of 24 letters}] & \text{Y} & [\beta \text{ of }24-\alpha\text{ letters}] & X & [\gamma \text{ of }24-\alpha-\beta\text{ letters}]\end{matrix}$$
I think the number of permutations of the letters surrounding Y and X will then be
$$\sum_{\alpha=0}^{24} \alpha! \binom{24}\alpha \sum_{\beta=0}^{24-\alpha} \beta! \binom{24-\alpha}\beta \sum_{\gamma=0}^{24-\alpha-\beta} \gamma! \binom{24-\alpha-\beta}\gamma = 506\,809\,361\,586\,340\,667\,433\,417\,385$$
By extension, an arrangement in which the vowels are alphabetized might look like
$$\begin{matrix}[\alpha \text{ consonants}] & \text{A} & [\beta \text{ consonants}] & \text{E} & [\gamma \text{ consonants}] & \text{I} & [\delta \text{ consonants}] & \text{O} & [\epsilon \text{ consonants}] & \text{U} & [\phi \text{ consonants}]\end{matrix}$$
with a total count of
$$\sum_{\alpha=0}^{21} \alpha! \binom{21}\alpha \sum_{\beta=0}^{21-\alpha} \beta! \binom{21-\alpha}\beta \sum_{\gamma=0}^{21-\alpha-\beta} \gamma! \binom{21-\alpha-\beta}\gamma \sum_{\delta=0}^{21-\alpha-\beta-\gamma} \delta! \binom{21-\alpha-\beta-\gamma}\delta \sum_{\epsilon=0}^{21-\alpha-\beta-\gamma-\delta} \epsilon! \binom{21-\alpha-\beta-\gamma-\delta}\epsilon \sum_{\phi=0}^{21-\alpha-\beta-\gamma-\delta-\epsilon} \phi! \binom{21-\alpha-\beta-\gamma-\delta-\epsilon}\phi = 11\,369\,881\,057\,625\,718\,121\,105$$
Am I counting these individual arrangements correctly?
Is there a more elegant way of counting them?
Possible solution
In the sum for the second count above, replace $21$ with $19$. Then multiply the resulting sum by the number of ways one can tack on Y and X to the arrangement. For a given alphabetical-vowel arrangement, Y can be placed in $25$ different positions. If we place Y at the $k$-th position $(1\le k\le25)$, then that leaves $26-k$ possible positions for X. The total number of ways to do this would be
$$25 \sum_{k=1}^{25} (26-k) = 8\,125$$
Then the total number of arrangements satisfying both criteria is
$$92\,380\,283\,593\,208\,959\,733\,978\,125$$
Is this count correct?
Best Answer
The total number of permutations is $26!$. Those where X follows Y are exactly half of them.
Now consider two permutations where X follows Y equivalent if they only differ for the position of the vowels. This defines an equivalence relation in which each class has precisely $5!$ elements and only one permutation in each equivalence class has the vowels in alphabetical order.
Thus the total number is $$ \frac{1}{2}\frac{26!}{5!}=1\,680\,381\,088\,027\,523\,481\,600\,000 $$
You get the same result if you use stars and bars. You want first to distribute the $21$ consonants (in alphabetical order) in six bins, which can well be empty: some (maybe zero) consonants before a, some (maybe zero) between a and e and so on.
The total number of possible distributions is $$ \binom{21+6-1}{6-1}=\binom{26}{5}=\frac{26!}{5!\,21!} $$ Now in any distribution you can permute the consonants freely in $21!$ ways, but at the end only half of them will have X after Y. Thus you get $$ \frac{1}{2}\frac{26!}{5!\,21!}21! $$ as before.