1. ## combinatory

How many different rearrangements of the word "facetious", has the letters a,e,i,o,u in alphabetical order.
I would like to know how to solve this problem.

2. ## Re: combinatory

Originally Posted by Yeison
How many different rearrangements of the word "facetious", has the letters a,e,i,o,u in alphabetical order.
I would like to know how to solve this problem.
There are $9!$ ways to rearrange the string "facetious".

But in each of those $aeiou$ has only one arrangement in alphabetical order.

So divide.

3. ## Re: combinatory

Originally Posted by Plato
There are $9!$ ways to rearrange the string "facetious".

But in each of those $aeiou$ has only one arrangement in alphabetical order.

So divide.
It took me a bit to digest this so let me add my two cents.

As Plato said there are 9! arrangements of facetious. This can be partitioned in $9 \cdot 8 \cdot 7 \cdot 6$ groups of $5!$

The first product represents the various ways your 4 consonants can be assigned to 9 slots.
Then 5! represents all the ways that the vowels can be permuted into the remaining slots.

However as Plato again noted only one of these 5! is in alphabetical order.

Spoiler:
So the the number of valid sequences is $\dfrac{9!}{5!}=9 \cdot 8 \cdot 7 \cdot 6$

4. ## Re: combinatory

Here's another 2 cents worth. I think counting problems are easy only after you have seen (or discovered) a solution and understand it.

For the problem at hand, first choose the 4 positions out of the 9 for the consonants. There are ${9\choose4}={9!\over5!4!}$ ways of doing this. For each such position, there are $4!$ ways of permuting the consonants. Thus there are ${9\choose4}\cdot4!={9!\over5!}$ total ways of placing the consonants. Once the consonants have been placed, there is only one way to place the vowels in alphabetical order. So the answer is ${9!\over5!}$.