Results 1 to 10 of 10

Math Help - Combinatorics Problem

  1. #1
    Junior Member
    Joined
    Aug 2007
    Posts
    30

    Combinatorics Problem

    In how many ways can 12 golf balls be distributed to 4 golfers in each of the following cases?

    b. The balls are different
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Think of the golfers as the boxes. So, to distribute r distinct objects into n different boxes you can use n^{r}

    4^{12}=16,777,216
    Last edited by galactus; February 12th 2008 at 07:11 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    3 balls each.

    so suppose we give them out in order

    we have \binom{12}{3} possibilities of balls for the first guy
    then \binom{9}{3} possibilities of balls for the second guy
    then \binom{6}{3} for the third
    and whatever is left over for the last guy

    this gives  \binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3} ways

    but we have not acounted for the order in which we distributed the balls so each distrubtion of the balls is repeated 4! times.

    giving a final answer of \frac{\binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3}}{4!} = 15400
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2007
    Posts
    30
    Bobak,

    yours only takes into account that each golfer can only have 3. Not including the cases in which one golfer has 6, one 5, one 1, and the other none.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    my bad I assumed it was an even distribution.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2007
    Posts
    30
    If it was evenly distributed it would be

    12!
    3!3!3!3!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,957
    Thanks
    1780
    Awards
    1
    If the instructions say that some of the golfers may receive no balls, then galactus has given you the answer.
    However, if we require that each golfer receives at least one ball the we are counting surjections: \mbox{surj}(12,4) = \sum\limits_{j = 0}^4 {( - 1)^j {4 \choose j}(4 - j)^{12} } .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Yes, since it wasn't stated I was assuming no restrictions.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    lerobert
    Guest

    Similar problem

    And what would be the solution if the balls are indistinctive?

    Namely, I have a similar problem. In how many ways can one put n blank sheets in k drawers, whereas a drawer can be empty?

    e.g. 100 sheets, 3 drawers
    d1 d2 d3
    100 0 0
    0 100 0
    23 45 32
    etc.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,957
    Thanks
    1780
    Awards
    1
    \binom {n+k-1} {n} is the number of ways to put n identical objects into k different cells.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. problem in combinatorics
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 21st 2011, 09:23 AM
  2. Help with combinatorics problem
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: December 7th 2010, 04:35 PM
  3. Combinatorics problem
    Posted in the Statistics Forum
    Replies: 6
    Last Post: November 12th 2009, 08:05 AM
  4. Combinatorics problem
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: August 14th 2009, 11:48 AM
  5. Combinatorics Problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 30th 2008, 06:11 AM

Search Tags


/mathhelpforum @mathhelpforum