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?


Do you think the book is incorrect?

I provide a third way to answer your question. First, arrange the vowels in alphabetical order and put a space between them and in front and back.
_a _e _e _e _e _i _o _u _
This leaves us 9 _'s  to put the 10 consonants rcrrncrltn in.
If we think of those blanks as being a variable we see that we are really only solving the diophantine equation
$\text{1) }a+b+c+d+e+f+g+h+i=10 , 10\geq a,b,c,d,e,f,g,h,i\geq0$
Where one possible answer to this 10+0+0+0+0+0+0+0+0 would be equivalent to arrangement rcrrncrltnaeeeeiou. So we see that the number of solutions to 1) is the number of ways we can stuff those consonants rcrrncrltn in between those ordered vowels.
To get the number of solutions we turn to generating functions. The generating function for 1) is
$\left(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}\right)^9$
We can expand this using Wolfram Alpha or any CAS to get:
$1 + 9 x + 45 x^2 + 165 x^3 + 495 x^4 + 1287 x^5 + 3003 x^6 + 
 6435 x^7 + 12870 x^8 + 24310 x^9 + 43758 x^{10} +...+$
We pick the coefficient of $x^{10}$ which is 43758. That is the number of ways to fit rcrrncrltn in between and around those ordered vowels. But we do not only want the consonants in the order of rcrrncrltn, the consonants can be permuted and to calculate that we just use a simple $\frac{10!}{4!2!2!}$
So the answer to your question is:
$\frac{43758 \cdot 10!}{4! 2! 2!}=1654052400$
which agrees with the other two answers. So, I would say the book answer is incorrect.
