How many arrangements of the letters of RECURRENCERELATION have the vowels in alphabetical order? This is my attempt at it.
We first arrange the vowels in alphabetical order:
A E E E E I O U
The following wedges indicate where we can place consonants
^ A ^ E ^ E ^ E ^ E ^ I ^ O ^ U ^ 
There can be more than one letter at each wedge. It is also possible that there are no letters at each wedge. 
There are $\frac{18!}{10!8!}$ ways to pick which wedges have consonants. Then there are $\frac{10!}{4!2!2!}$ ways to arrange the letters within the choosen spots. Thus my final answer is $\frac{18!}{10!8!}\frac{10!}{4!2!2!}$ ways.
I think my problem is how I counted the number of ways to place letters in the wedges. Can anyone help me fix it?

Bear with me.
Suppose there are $K$ ways to arrange the vowels.  Then for any arrangement of letters there will be $K$ equivalent ways that have the exact same consonants in the exact same possitions but with the vowells in possibly different places.  So for example there will be $K$ possible ways to make words of the form R-C-RR-NC-R-L-T--N where - s are vowels.
This means if there is one way to arrange the vowels in alphabetical order and there are $N$ ways to arrange the letters in any order then there are $\frac NK$ ways to arrange the letters with the vowels in alphabetical order.
So what is $K$ and what is $N$.
If ordered matter there would be $8!$ ways to arrange the $8$ vowels. But as there are $4$ Es there are $\frac {8!}{4!}$ ways to do it.  $K =\frac {8!}{4!}$.
Likewise $N = \frac{18!}{4!4!2!2!}$ as there are $18$ letters and $4$ Rs, $4$ Es, $2$ Cs, and $2$ Ns.
So the answer is $\frac {N}{K} = \frac { \frac{18!}{4!4!2!2!}}{\frac {8!}{4!}} = \frac {18!4!}{4!4!2!2!8!}= \frac{18!}{4!2!2!8!}$
I think.  I hope.
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Okay, inefficient but thouroug second way.
If the $R$s were all different there'd be $18*17*16*15$ slots to place them.  But there are $4$ $R$s so there are $\frac{18*17*16*15}{4!} = \frac {18!}{14!4!} = {18 \choose 4}$ to place the $R$.  There are ${14 \choose 2}$ ways to place the $2$ $N$s. And ${12 \choose 2}$ ways to place the $C$s.  Then $10*9$ ways to place the $T$ and $L$.  That leaves the vowels.  And the must be in alphabetical order.  So:
$\frac{18!}{14!4!}*\frac{14!}{12!2!}*\frac{12!}{10!2!}*10*9$
$=\frac{18!}{4!2!2!8!}$.
Hmm.  Don't see how the book can be right.  There are 18 places to put the first letter and 17 places to put the second.  So 17 must be a factor even if I screwed everything else up.  The book's answer has no prime factors greater than 11$.
