Results 1 to 6 of 6

Math Help - Permutation

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    60

    Permutation

    Hi, just to make sure I got the right idea.

    In how many ways can the letters of the word MERETRICIOUS be arranged in a line?

    Since there are 2E, 2I and 2R. Would the first answer be 12! / (2! 2! 2!)?

    What is the probability that an arrangement begins with M and ends with S.

    Would the probability be (10!/(2! 2! 2!) / (12!/ (2! 2! 2!) as a result?

    Thanks for any clarification.
    Last edited by Geometor; October 12th 2008 at 07:22 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2008
    Posts
    46
    The first part is right.

    I don't think the second part is right though. The probability of choosing an M as the first letter is going to be 1choose(12) = n!/(n-1)!k!
    So,

    1choose(12) = 1!/11!12!

    Now we want to choose an S as the last letter. The probability is now 1choose11 since we have already chosen an M and there are 1 fewer letters to choose. So,

    1choose(11) = 1!/10!11!

    Since we want both of these events to happen we multiply them together. To get the probability of this happening we divide by the total number of possibilities which you have calculated in the first part. So,

    (1choose(12) * 1choose(11)) / (12!/(2!2!2!))

    Hopefully this makes sense.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2007
    Posts
    60
    Choose?
    I'm abit confused since I believe choose refers to combinations rather than permutations. And I think permutation are needed in this case since the order of arrangement does count as well

    ah well, thanks for helping anyway
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,957
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by Geometor View Post
    I'm abit confused since I believe choose refers to combinations rather than permutations. And I think permutation are needed in this case since the order of arrangement does count as well
    Actually you first answer is correct.
    \frac{{\frac{{10!}}<br />
{{\left( 2 \right)^3 }}}}<br />
{{\frac{{12!}}<br />
{{\left( 2 \right)^3 }}}} = \frac{1}<br />
{{12\left( {11} \right)}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    917
    Thanks
    27

    Permutatons

    originally posted by geometer

    I agree with your solution with one exception.Shoudn't the numerator in the answer equation be 10 factorial divided by 16 to account for the fact that m or s can head the string requiring halving again


    bj
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Geometor View Post
    Hi, just to make sure I got the right idea.

    In how many ways can the letters of the word MERETRICIOUS be arranged in a line?

    Since there are 2E, 2I and 2R. Would the first answer be 12! / (2! 2! 2!)?

    What is the probability that an arrangement begins with M and ends with S.

    Would the probability be (10!/(2! 2! 2!) / (12!/ (2! 2! 2!) as a result?

    Thanks for any clarification.
    All your calculations are correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Why permutation ?
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: July 11th 2010, 12:27 PM
  2. permutation help
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: July 9th 2010, 03:37 PM
  3. Permutation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: August 7th 2008, 12:01 PM
  4. Permutation.......
    Posted in the Statistics Forum
    Replies: 5
    Last Post: March 24th 2008, 04:41 AM
  5. not a permutation? then how?
    Posted in the Statistics Forum
    Replies: 3
    Last Post: March 16th 2008, 01:50 AM

Search Tags


/mathhelpforum @mathhelpforum