Results 1 to 3 of 3

Math Help - combinations problem

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    84

    combinations problem

    ok, so i'm having trouble with the following problem, so if anyone could help, i'd be grateful. i think that there's some kind of permutation needed for this (meaning n choose k), but i'm confused about how to use it.

    A bowling team has 11 adults and 7 children. One night, they decide to divide themselves into 3 teams. The first will have 3 adults and 3 children. The second will have 4 adults and 2 children. The third will also have 4 adults and 2 children. In how many ways can the three teams be chosen?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, clockingly!

    I think you mean "combinations".
    The order of the names on a team is not considered.


    A bowling team has 11 adults and 7 children.
    One night, they decide to divide themselves into 3 teams.
    . . The first will have 3 adults and 3 children.
    . . The second will have 4 adults and 2 children.
    . . The third will also have 4 adults and 2 children.

    In how many ways can the three teams be chosen?

    I will assume that the three teams are ordered and distinct.
    That is, that they are called "Team #1", "Team #2", and "Team #3". **


    Select team #1.
    There are 11 adults; we choose 3 of them: \binom{11}{3} ways.
    There are 7 children; we choose 3 of them: \binom{7}{3} ways.
    Hence, there are: . \binom{11}{3}\binom{7}{3}\:=\:165\cdot35 \:=\:5,775 ways to select team #1.


    Select team #2.
    There are 8 available adults; we choose 4 of them: \binom{8}{4} ways.
    There are 4 available children; we choose 2 of them: \binom{4}{2} ways.
    Hence, there are: . \binom{8}{4}\binom{4}{2} \:=\:70\cdot6\:=\:420 ways to select team #2.

    The remaining 4 adults and 2 children will belong to team #3.


    Therefore, there are: . 5,775 \times 420 \:=\:\boxed{2,425,500} ways to select the teams.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** If teams #2 and #3 are interchangeable,
    . . . then the "420" is twice as large as it should be.

    The number of ways would be: . 5,775 \times 210 \:=\:\boxed{1,212,750}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2006
    Posts
    84
    Thank you so much! You explained that in a very clear way.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations problem
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: October 14th 2011, 04:52 PM
  2. Plz Help - combinations problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 12th 2009, 11:29 AM
  3. Combinations problem
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 21st 2009, 08:42 AM
  4. Combinations problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: April 26th 2009, 08:48 PM
  5. combinations problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: February 3rd 2009, 02:10 PM

Search Tags


/mathhelpforum @mathhelpforum