Difference between revisions of "Ball-and-urn"

(restored to January 27)
Line 19: Line 19:
== Problems ==
== Problems ==
*[[2003 AMC 10A Problems/Problem 21]]
*[[2003 AMC 10A Problems/Problem 21]]
*[[2006 AMC 10B Problems/Problem 17]]
*[[Mock AIME 3 Pre 2005 Problems/Problem 2]]
*[[Mock AIME 3 Pre 2005 Problems/Problem 2]]
*[[2007 AIME I Problems/Problem 10]]
*[[2007 AIME I Problems/Problem 10]]

Revision as of 10:26, 10 March 2018

The ball-and-urn technique, also known as stars-and-bars, is a commonly used technique in combinatorics.

It is used to solve problems of the form: how many ways can one distribute $k$ indistinguishable objects into $n$ bins? We can imagine this as finding the number of ways to drop $k$ balls into $n$ urns, or equivalently to drop $k$ balls amongst $n-1$ dividers. The number of ways to do such is ${n+k-1 \choose k}$.

Reasoning (One of Several)

If you could only put one ball in each urn, then there would be ${n \choose k}$ possibilities; the problem is that you can repeat urns, so this does not work. You can, however, reframe the problem as so: imagine that you have the $n$ urns (numbered 1 through $n$) and then you also have $k-1$ urns labeled "repeat 1st", "repeat 2nd", ..., "repeat $k-1$st". After the balls are in urns you can imagine that any balls in the "repeat" urns are moved on top of the correct balls in the first $n$ urns, moving from left to right. There is a one-to-one correspondence between the non-repeating arrangements in these new urns and the repeats-allowed arrangements in the original urns.

For a simple example, consider 4 balls and 5 urns. The one to one correspondence between several of the possibilities and the "repeated urns" version is shown. Since there are 4 balls, these examples will have three possible "repeat" urns. For simplicity, I am listing the numbers of the urns with balls in them, so "1,1,2,4" means 2 balls in urn 1, 1 in urn 2, and 1 in urn 4. The same is true for the "repeat" urns options but I use the notation $r_1$ etc.

  • 1,2,3,4 <-> 1,2,3,4 (no repeats)
  • 1,1,2,4 <-> 1,2,4, $r_1$
  • 1,2,2,2 <-> 1,2, $r_2$, $r_3$
  • 4,4,5,5 <-> 4,5, $r_1$, $r_2$

Since the reframed version of the problem has n+k-1 urns, and k balls that can each only go in one urn, the number of possible scenarios is simply ${n+k-1  \choose k}$


Invalid username
Login to AoPS