Results 1 to 4 of 4
Like Tree3Thanks
  • 1 Post By romsek
  • 1 Post By Plato
  • 1 Post By Soroban

Math Help - Combinatory

  1. #1
    Junior Member
    Joined
    Feb 2014
    From
    USA
    Posts
    28
    Thanks
    3

    Combinatory

    I have a lot of doubts on how to solve this problem.
    How many different ways can I put 12 books into 4 different bookshelves if all the books are different and the position of the books in each bookshelf is important.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,355
    Thanks
    908

    Re: Combinatory

    Quote Originally Posted by Yeison View Post
    I have a lot of doubts on how to solve this problem.
    How many different ways can I put 12 books into 4 different bookshelves if all the books are different and the position of the books in each bookshelf is important.
    I always end up getting these frigging combinatorics problems wrong but what I think the answer here is is

    $12! 4^{12}$

    You first select a book from the 12 and put it on one of 4 shelves. There are 12x4 ways to do this.
    Then you select a book from the 11 and put it on one of 4 shelves. There are 11x4 ways to do this.
    And so on. The final product is

    $(12\cdot 4)(11 \cdot 4) \dots (1 \cdot 4) = 12! 4^{12}$
    Thanks from Yeison
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1

    Re: Combinatory

    Quote Originally Posted by romsek View Post
    $12! 4^{12}$
    That is correct if we assume that each book shelf can hold all twelve books.
    Thanks from Yeison
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    632

    Re: Combinatory

    Hello, romsek!


    You first select a book from the 12 and put it on one of 4 shelves.
    . . There are 12x4 ways to do this.

    Then you select a book from the 11 and put it on one of 4 shelves.

    And place it to the right of the previous book.
    . . There are 11x4 ways to do this.

    And so on.

    The final product is: .$(12\cdot 4)(11 \cdot 4) \dots (1 \cdot 4) = 12! \:\!4^{12}$

    Brilliant!

    Since you arrived last November,
    . . you have provided many elegant solutions.

    A belated "Welcome board", Romsek!
    Thanks from Yeison
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinatory
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 17th 2014, 03:14 PM
  2. Combinatory
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: March 3rd 2014, 06:48 PM
  3. combinatory
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: March 2nd 2014, 08:54 AM
  4. Combinatory Counting help
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: February 14th 2009, 09:07 AM
  5. Combinatory problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 28th 2008, 05:36 PM

Search Tags


/mathhelpforum @mathhelpforum