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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?

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You don't want to place consonants between vowels as though you we jumbling balls into boxes, you want to place vowels among consonants in a specific order.   Use identical placeholders.

Consider, that the task can be performed by rearranging the letters of $\rm R\color{blue}XC\color{blue}XRR\color{blue}XNC\color{blue}XR\color{blue}XL\color{blue}XT\color{blue}{XX}N$, then replacing the $\rm\color{blue}X$ with the vowels in alphabetical order ($\rm AEEEEIOU$).

How many distinct ways are there to rearrange 18 literals consisting of 8 X, 4 R, 2 N, 2 C, 1 T, 1 L?

$$\dfrac{18!}{8! 4! 2! 2!}$$

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  • $\begingroup$ Thanks, I agree with your answer. The answer in the back of the book is $\frac{10!}{4!2!2!}\binom{11}{8}$. Do you think it is incorrect? $\endgroup$ – user100000000000000 Feb 9 '17 at 2:33
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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$.

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  • $\begingroup$ Your answer makes sense to me. But the answer in the back of the book is $\frac{10!}{4!2!2!}\binom{11}{8}$. Do you think the book is incorrect? $\endgroup$ – user100000000000000 Feb 9 '17 at 3:08
  • $\begingroup$ I could have made an error. Not sure though. The book could be wrong but I could be too. $\endgroup$ – fleablood Feb 9 '17 at 4:07
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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.

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