Results 1 to 5 of 5

Math Help - Permutation of items with a constant order subset.

  1. #1
    Newbie driegert's Avatar
    Joined
    Feb 2010
    From
    Kingston, Ontario
    Posts
    16

    Permutation of items with a constant order subset.

    I just wrote a practice midterm for a probability course. There was no answer key and the person who "marked" the exam and I were unsure about one of the questions.

    "What is the probability that in a random permutation of the nine letters A, E, I, O, U, C, F, S, and T, the vowels appear in alphabetic order -- A, E, I, O, U (e.g., as in "FACETIOUS")?"

    I ended up with:

    \begin{align*}P(A) &= \frac{\binom{4}{1} \cdot 6 \cdot \binom{3}{1} \cdot 7 \cdot \binom{2}{1} \cdot 8 \cdot \binom{1}{1} \cdot 9}{9!} \\ &= \frac{(9 \cdot 8 \cdot 7 \cdot 6) \cdot 4!}{9!} \\ &= 0.2 \end{align*}

    My rational being that you order A, E, I, O, U and you have 6 choices of where to put the first of the remaining 4 letters: _ A _ E _ I _ O _ U _ . So, choose one of those letters, \binom{4}{1} and multiply by the 6 possibilities. Once the first letter is placed, it opens up a new spot. Choose the next letter, \binom{3}{1}, and place it in one of the 7 remaining spots. (Same for last two remaining letters).

    The marker ended up with:

    \begin{align*} P(A) &= \frac{\binom{9}{5}4!}{9!} \\ &= \frac{9 \cdot 8 \cdot 7 \cdot 6}{9!} \\ &= 0.0083 \bar{3} \end{align*}

    His rational being that you have 9 spots to fill, choose 5 of them for the vowels, and arrange the remaining four letters in the remaining 4 spots.

    I ended up with an additional 4!, but neither of us could see where I (most likely) went wrong.

    Any clarification you could provide would be greatly appreciated .

    Thanks,
    --
    Dave

    (Hopefully I explained that properly.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1

    Re: Permutation of items with a constant order subset.

    I would calculate, using the fact that there are \binom{11}{5}
    ways to have A, E, I, O, U in that order.

    For each of these, there are 6! ways to arrange the remaining letters.

    Hence, the probability is

    \frac{6!\binom{11}{5}}{11!}=0.008333...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,540
    Thanks
    780

    Re: Permutation of items with a constant order subset.

    The marker is right.

    Quote Originally Posted by driegert View Post
    My rational being that you order A, E, I, O, U and you have 6 choices of where to put the first of the remaining 4 letters: _ A _ E _ I _ O _ U _ . So, choose one of those letters, \binom{4}{1} and multiply by the 6 possibilities. Once the first letter is placed, it opens up a new spot. Choose the next letter, \binom{3}{1}, and place it in one of the 7 remaining spots. (Same for last two remaining letters).
    Suppose at first you chose F and decided to insert it between A and E. Then suppose you chose C and inserted it between F and E. On the other hand, you could choose C first and insert it between A and E, then choose F and insert it between A and C. In both cases, you end up with _ A _F _ C _ E _ I _ O _ U _. Thus, you count several variants twice.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie driegert's Avatar
    Joined
    Feb 2010
    From
    Kingston, Ontario
    Posts
    16

    Re: Permutation of items with a constant order subset.

    Thanks! I was thinking I had double permuted the remaining letters.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,708
    Thanks
    1638
    Awards
    1

    Re: Permutation of items with a constant order subset.

    Quote Originally Posted by driegert View Post
    "What is the probability that in a random permutation of the nine letters A, E, I, O, U, C, F, S, and T, the vowels appear in alphabetic order -- A, E, I, O, U (e.g., as in "FACETIOUS")?"
    There is a simple way to look at this problem.
    The answer is simply \frac{1}{5!}=\frac{1}{120}=0.008\overline{3}

    Look at it this way. Only 1 in every 120 arrangements do the vowels appear in alphabetic order.
    It is the same answer if the set were
    \{A,B,C,E,I,K,L,M,N,O,U,X,Y,Z\}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. order of a permutation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 29th 2010, 12:07 AM
  2. Order of items around a circle
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 28th 2010, 10:52 AM
  3. Replies: 4
    Last Post: September 18th 2010, 07:41 PM
  4. order of permutation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 19th 2008, 07:41 PM
  5. [SOLVED] Order of Permutation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 18th 2006, 05:46 PM

Search Tags


/mathhelpforum @mathhelpforum