Results 1 to 8 of 8

Math Help - Permutations problem

  1. #1
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419

    Permutations problem

    Not really probability/statistics but this is an introductory discrete mathematics class and I'm rusty on all of this material, so here it goes:

    Pamela has 15 different books. In how many ways can she place her books on two shelves so that there is at least one book on each shelf?

    So I considered the one event where 14 books are on Shelf A and only 1 on Shelf B, and so the possible arrangement of books in that scenario is 14! \times 1!. Now, if there are 13 books on Shelf A and 2 are on Shelf B, the number of possible arrangements are 13! \times 2!. And so it goes on as such and so the total possible arrangements would be 2(14! \times 1! + 13!\times 2! + ... + 9! \times 6! + 8! \times 7!). Is this correct or am I just way off? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by Pinkk View Post
    Not really probability/statistics but this is an introductory discrete mathematics class and I'm rusty on all of this material, so here it goes:

    Pamela has 15 different books. In how many ways can she place her books on two shelves so that there is at least one book on each shelf?

    So I considered the one event where 14 books are on Shelf A and only 1 on Shelf B, and so the possible arrangement of books in that scenario is 14! \times 1!. Now, if there are 13 books on Shelf A and 2 are on Shelf B, the number of possible arrangements are 13! \times 2!. And so it goes on as such and so the total possible arrangements would be 2(14! \times 1! + 13!\times 2! + ... + 9! \times 6! + 8! \times 7!). Is this correct or am I just way off? Thanks.
    Hi Pinkk,

    In the case where there are 8 books on the first shelf and 7 books on the second, for example, have you accounted for the number of ways you can split the books into a set of 8 and a set of 7?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    Isn't that taken into account with the 2 in front of everything in the parentheses? Because you can have 8 on shelf A and 7 on shelf B, or 7 on shelf A and 8 on shelf B, and in either case, the way you can arrange 8 on one shelf is 8! and the way you can arrange 7 on another shelf is 7!, so the way you can arrange 8 on one shelf and 7 on another is 8! \times 7!.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Pinkk View Post
    Isn't that taken into account with the 2 in front of everything in the parentheses? Because you can have 8 on shelf A and 7 on shelf B, or 7 on shelf A and 8 on shelf B, and in either case, the way you can arrange 8 on one shelf is 8! and the way you can arrange 7 on another shelf is 7!, so the way you can arrange 8 on one shelf and 7 on another is 8! \times 7!.
    You can arrange the 8 books with 8! on one shelf and the 7 on the other shelf as 7!
    but that only takes into account the specific 8 books chosen from the 15
    and the remaining 7 books.
    You can split the 15 into 8 and 7 in many different ways, specifically \binom{15}{8}

    You appear to want to count all possible arrangements, rather than just groups.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, Pinkk!

    Pamela has 15 different books.
    In how many ways can she place her books on two shelves
    so that there is at least one book on each shelf?

    For each of the 15 books, Pamela makes a decision:
    . . (1) Place it on shelf A, or (3) place it on shelf B.

    There are: . 2^{15} \:=\:32,768 possible choices she can make.


    However, these include these two disallowed cases:
    . . All 15 books are on shelf A, and all 15 books are on shelf B.


    Therefore, there are: . 32,768 - 2 \:=\:32,766 ways to place the books
    . . so that there is at least one book on each shelf.

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1576
    Awards
    1
    The last reply to this thread is beyond me.
    Let us we understand the question as:
    “How many ways can we arrange fifteen books on two shelves so that no self is empty”
    Then there are 15! ways to arrange those books.
    For any one of those ways, there are fourteen ways to place that string onto two slelves.
    Thus 14\cdot 15! =18307441152000
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    May 2010
    Posts
    13
    I read it like the question and answer like this --

    If order does not matter then the question is like a coin flip and yes it would be 2^{15} -2, however order is likely to matter.

    Because order does matter I think it would be 15! \cdot 14 = 1.6999 E^{13}

    As a workable example if you had three books and order mattered there would be 24 ways to split across the two shelves.

    6 ways for 0 books on the top shelf (/abc, /acb, /bac, /bca, /cab, /cba)
    6 ways for 1 book on the top shelf (a/bc, a/cb, b/ac, b/ca, c/ab, c/ba)
    6 ways for 2 books on the top shelf (ab/c, ba/c, ac/b, ca/b, bc/a, cb/a)
    6 ways for 3 books on the top shelf (abc/, acb/, bac/, bca/, cab/, cba/)

    So it's going to be n! books * ways to split. Because we can't have 0 or 15 books on the top shelf there are 14 ways (1 - 14) to split the books.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, Plato!

    You're right . . . I forgot about the ordering of the books.
    (There are 15 different books . . . I knew that! . . . *blush*)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] permutations problem
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: April 19th 2010, 09:29 AM
  2. tricky permutations problem
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: March 31st 2010, 02:32 PM
  3. Permutations and Combinations problem
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 3rd 2009, 08:17 PM
  4. permutations problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: September 12th 2008, 04:45 PM
  5. Permutations and Combinations Problem!! Help!
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 24th 2008, 04:35 AM

Search Tags


/mathhelpforum @mathhelpforum