Results 1 to 11 of 11

Math Help - Fundamental Counting Principles

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    23

    Fundamental Counting Principles

    20 players are to be divided into two 10-man teams. In how many ways can that be done?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kturf View Post
    20 players are to be divided into two 10-man teams. In how many ways can that be done?
    What do you think?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    23

    Not sure but ...

    Here is a thought:
    20!/10!(20-10)!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,786
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by kturf View Post
    20 players are to be divided into two 10-man teams. In how many ways can that be done?
    These are an unordered partitions.
    We can divide thirty people in five groups of six each in \frac{30!}{(6!)^5(5!)}ways.
    So what is the answer to your question?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    13
    I see it so. The sequence is not important, so
    {20 \choose 10}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2010
    Posts
    13
    Quote Originally Posted by Plato View Post
    These are an unordered partitions.
    We can divide thirty people in five groups of six each in \frac{30!}{(6!)^5(5!)}ways.

    Could you show why?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,786
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by Oloria View Post
    I see it so. The sequence is not important, so a group: a, b is the same as a group: b, a.
    {10 \choose 2}{8 \choose 2}{6 \choose 2}{4 \choose 2}{2 \choose 2}
    That is not correct. Read my first reply.
    There are only two teams of ten each. The teams are un-named.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2010
    Posts
    13
    I noticed it on my own before. I read too quickly. But I asked you about something another...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,786
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by Oloria View Post
    Could you show why?
    Say we have a prep school class of fifteen boys.
    We partition them into three study groups of five each to study: biology, mathematics, and geography.
    This can be done in \binom{15}{5}\binom{10}{5} \binom{5}{5}=\frac{15!}{(5!)^3} ways.
    That is an ordered partition because names of the groups as well as the content of the of the group matters.

    NOW, We partition them into three teams of five each to play.
    This can be done in \frac{15!}{(5!)^3(3!)} ways.
    That is an unordered partition because the teams are not named so only the content of the of the team matters.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2009
    Posts
    23

    OK - that helps.

    I think the answer is:

    20!/(10!)^2(2!)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,786
    Thanks
    1683
    Awards
    1
    Quote Originally Posted by kturf View Post
    I think the answer is:

    20!/(10!)^2(2!)
    Correct!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basic Counting Principles
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 23rd 2010, 09:35 AM
  2. Counting Principles
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: May 23rd 2010, 12:18 AM
  3. [SOLVED] Counting principles
    Posted in the Statistics Forum
    Replies: 2
    Last Post: April 3rd 2009, 02:13 PM
  4. Counting Principles
    Posted in the Statistics Forum
    Replies: 3
    Last Post: April 2nd 2008, 08:23 PM
  5. counting principles
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 26th 2007, 06:06 AM

Search Tags


/mathhelpforum @mathhelpforum