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

• Sep 12th 2012, 02:08 PM
light1985
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.
• Sep 12th 2012, 02:21 PM
Plato
Re: Counting problem: Placing identical balls into unique bins with capacity constrai
Quote:

Originally Posted by light1985
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?
• Sep 12th 2012, 04:37 PM
light1985
Re: Counting problem: Placing identical balls into unique bins with capacity constrai
Quote:

Originally Posted by Plato
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.