# How many arrangements of the letters in COMBINATORICS have no consecutive vowels?

I have a problem here that I've been mulling over for a while and I can't seem to grasp why the book has a different answer than me.

Here's my attempted solution -

As per "Applied Combinatorics" I noted that it's a "consecutive vowel" problem and immediately went to creating a diagram that looks like below

$$\_v\_v\_v\_v\_v\_$$

Where each _ is a "box". Now, I know there are 5 vowels and 8 consonants, hence the 5 v's.

By doing this I reduced the problem to a "distribute n objects into r boxes with repetition" problem and concluded with the answer

$$\binom{4+6-1}{4}6!5!$$

Which, I suspect, would be the answer if there weren't repeated letters. However, this is wrong and the book lists the answer as:

$$\binom{9}{4} * \frac{8!}{2!} * \frac{5!}{2!2!}$$

Given that the Cs repeat, as well as the O's and I's I'm at a loss as to how to handle this problem (mostly because the example my textbook gives is so trivial it's hard to see beyond it).

Can anyone clear this problem up for me? Thank you!

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## 1 Answer

First, arrange the 8 consonants (with one repeat of a double C) in $\frac{8!}{2!}$ distinct ways.

Now arrange the five vowels, with two repeats, double I, and double O, in $\frac{5!}{2! *2!}$ ways.

Finally, select 5 of the 9 slots for vowels, one at the start of the consonants, and one after each consonant.

Multiply. Done.

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Shouldn't that be 9 choose 5 rather than 9 choose 4 like the book suggests? I should note that this wouldn't be the first time the book was wrong. Some of the answers from the previous edition made it into this edition and didn't get caught. –  anonacct1405 Mar 11 '13 at 0:08
Uhh... What's the difference? Or, consider choosing the 4 slots to be empty... –  DJohnM Mar 11 '13 at 0:13