Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Counting problem: Placing identical balls into unique bins with capacity constraint

  1. #1
    Newbie
    Joined
    Sep 2012
    From
    United States
    Posts
    3

    Counting problem: Placing identical balls into unique bins with capacity constraint

    I recently encountered a real-life scenario where I need to enumerate all the possible combinations. The problem could be framed as follows:

    Suppose there are 14 identical balls and 5 unique bins that can hold 6 balls each. What are the total number of ways to place the 14 balls into the 5 bins?

    Using a divide and conquer strategy (basically I first look at what happens when there are only 2 unique bins, then 4 unique bins, and finally 5), I manage to come up with an upper bound of 3060. But I have no idea how to handle the constraint that each bin can only hold 6 balls.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1

    Re: Counting problem: Placing identical balls into unique bins with capacity constrai

    Quote Originally Posted by light1985 View Post
    I recently encountered a real-life scenario where I need to enumerate all the possible combinations. The problem could be framed as follows:
    Suppose there are 14 identical balls and 5 unique bins that can hold 6 balls each. What are the total number of ways to place the 14 balls into the 5 bins?.
    Look at this webpage.
    The coefficient of x^{14} is the actual answer.

    P.S Edit
    Here is a strictly Discrete Mathematics answer, in the "back-of-the-book" format.
    \binom{14+4}{4}-5\sum\limits_{k = 1}^7 {\binom{k+3}{3}}+5=1420 .

    Now WHY?
    Last edited by Plato; September 12th 2012 at 03:36 PM.
    Thanks from light1985
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2012
    From
    United States
    Posts
    3

    Re: Counting problem: Placing identical balls into unique bins with capacity constrai

    Quote Originally Posted by Plato View Post
    Look at this webpage.
    The coefficient of x^{14} is the actual answer.

    P.S Edit
    Here is a strictly Discrete Mathematics answer, in the "back-of-the-book" format.
    \binom{14+4}{4}-5\sum\limits_{k = 1}^7 {\binom{k+3}{3}}+5=1420 .

    Now WHY?
    Thanks, I figured out what each of the terms represents.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. n balls in d bins (non-standard?)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 18th 2012, 04:47 AM
  2. probability of balls falling to bins
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: January 20th 2011, 02:02 PM
  3. STATA - Placing a Constraint on a Constant
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 30th 2010, 02:11 PM
  4. different balls, identical boxes
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 21st 2010, 01:29 AM
  5. Placing Identical/Distinct Objects
    Posted in the Statistics Forum
    Replies: 4
    Last Post: May 12th 2009, 03:21 PM

Search Tags


/mathhelpforum @mathhelpforum