Results 1 to 5 of 5

Math Help - partioning sets...

  1. #1
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    partioning sets...

    a set has 5 members.what is the the number ways of partioning it into 2 or more subsets???
    please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17
    What things are you given and roughly what knowledge are you supposed to use?
    Last edited by Vlasev; August 12th 2010 at 12:41 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9
    all i have is the question.
    you should be able to answer it with only high school maths.
    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 earthboy View Post
    a set has 5 members.what is the the number ways of partioning it into 2 or more subsets???
    Quote Originally Posted by earthboy View Post
    all i have is the question.
    you should be able to answer it with only high school maths.
    I do not consider this to a high school level question.
    So I am giving you the technical answer. But I cannot explain to you.
    The fifth Bell number is \mathcal{B}(5)=52.
    That is the number of ways to partition a set of five.
    But one of those is the set itself, so it does meet the condition of ‘two or more subsets’.
    Thus you answer must be 51.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by earthboy View Post
    all i have is the question.
    you should be able to answer it with only high school maths.
    You could try it using n_C_r=\frac{n!}{r!(n-r)!}

    You cannot have more than 4 members in a subset.
    A subset must contain at least one member.

    Therefore you could list the possibilities, then count the number of ways those possibilities can occur.

    (a) 4 members in a subset and 1 member in the other subset

    For this option, you can count using combinations the number of ways to select 4 from 5 or 1 from 5.


    (b) (i) 3 members in a subset and 2 members in another subset

    You only need to count the number of ways of selecting 3 from 5 or 2 from 5.
    (Selecting any 3 automatically selects the remaining 2 and vice versa)

    (b) (ii) 3 members in a subset and 2 subsets containing 1 member

    You only need to count the number of ways to choose 3 from 5.


    (c) (i) 2 members in a subset and 2 members in another subset
    ..........(the 5th member automatically makes the 3rd subset of 1)


    Select 4 of the 5 and count the number of ways to form 2 groups of 2.
    Hence, first count the number of ways to select 4 from 5,
    then for each group of 4, count the number of ways to pair one member with one of the other 3.


    (c) (ii) 2 members in a subset and the other 3 making single-member subsets

    Select 2 from 5 as the remainder automatically form the single-member subsets.


    (d) One member per subset
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Open sets and sets of interior points
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 9th 2011, 04:10 AM
  2. Metric spaces, open sets, and closed sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 16th 2011, 06:17 PM
  3. Replies: 9
    Last Post: November 6th 2010, 01:47 PM
  4. Open sets and closed sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 30th 2010, 11:05 AM
  5. Approximation of borel sets from the top with closed sets.
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 18th 2010, 09:51 AM

Search Tags


/mathhelpforum @mathhelpforum