Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By romsek
  • 1 Post By johng

Math Help - combinatory

  1. #1
    Junior Member
    Joined
    Feb 2014
    From
    USA
    Posts
    33
    Thanks
    3

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,962
    Thanks
    1784
    Awards
    1

    Re: combinatory

    Quote Originally Posted by Yeison View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,789
    Thanks
    1147

    Re: combinatory

    Quote Originally Posted by Plato View Post
    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 $\dfrac{9!}{5!}=9 \cdot 8 \cdot 7 \cdot 6$
    Thanks from Yeison
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    709
    Thanks
    294

    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 ${9!\over5!}$.
    Thanks from Yeison
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinatory optimization problem
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: April 2nd 2010, 01:54 PM
  2. combinatory question
    Posted in the Discrete Math Forum
    Replies: 10
    Last Post: October 14th 2009, 09:13 AM
  3. Combinatory Counting help
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: February 14th 2009, 10:07 AM
  4. Combinatory problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 28th 2008, 06:36 PM

Search Tags


/mathhelpforum @mathhelpforum