Difference between revisions of "Ball-and-urn"

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The [[ball-and-urn]] technique, also known as [[stars-and-bars]], is a commonly used technique in [[combinatorics]].
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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 <math>k</math> indistinguishable objects into <math>n</math> bins? We can imagine this as finding the number of ways to drop <math>k</math> balls into <math>n</math> urns, or equivalently to drop <math>k</math> balls amongst <math>n-1</math> dividers. The number of ways to do such is <math>{n+k-1 \choose k}</math>.
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It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguishable objects into <math>n</math> distinguishable bins? We can imagine this as finding the number of ways to drop <math>k</math> balls into <math>n</math> urns, or equivalently to arrange <math>k</math> balls and <math>n-1</math> dividers. For example, <cmath>****||</cmath><cmath>***|*|</cmath><cmath>*|**|*</cmath><cmath>...</cmath> represent the ways to put <math>k=4</math> objects in <math>n=3</math> bins. The number of ways to do such is <math>{n+k-1 \choose k}</math>, or <math>{n+k-1 \choose n-1}</math>.
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== Reasoning (One of Several) ==
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Arranging <math>k</math> *'s and <math>n-1</math> |'s is the same as saying there are <math>k+n-1</math> positions: <cmath>\underbrace{\_~~ \_~~ \_~~ \_~~ \_~~ \_~~}_{k+n-1}</cmath> and you want to fill <math>k</math> of them with *'s and the rest of them with |'s. Thus you are choosing <math>k</math> positions out of <math>n+k-1</math> total positions, resulting in a total of <math>{n+k-1\choose k}</math> ways.
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== Reasoning (One of Several) ==
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If you could only put one ball in each urn, then there would be <math>{n \choose k}</math> 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 <math>n</math> urns (numbered 1 through <math>n</math>) and then you also have <math>k-1</math> urns labeled "repeat 1st", "repeat 2nd", ..., and "repeat <math>k-1</math>-th". 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 <math>n</math> 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.
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For a simple example, consider <math>4</math> balls and <math>5</math> 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 <math>2</math> balls in urn <math>1, 1</math> in urn <math>2,</math> and <math>1</math> in urn <math>4.</math> The same is true for the "repeat" urns options but I use the notation <math>r_1</math> etc.
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*<math>1,2,3,4 \leftrightarrow 1,2,3,4</math> (no repeats).
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*<math>1,1,2,4 \leftrightarrow 1,2,4,</math> <math>r_1</math>.
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*<math>1,2,2,2 \leftrightarrow 1,2,</math> <math>r_2</math>, <math>r_3</math>.
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*<math>4,4,5,5 \leftrightarrow 4,5,</math> <math>r_1</math>, <math>r_2</math>.
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Since the re-framed version of the problem has <math>n+k-1</math> urns, and <math>k</math> balls that can each only go in one urn, the number of possible scenarios is simply <math>{n+k-1
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\choose k}.</math> Note: Due to the principle that <math>{a \choose b}={a \choose a-b}</math>, we can say that <math>{n+k-1 \choose k}={n+k-1 \choose n-1}</math>.
  
 
== Problems ==
 
== Problems ==
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*[[1986 AIME Problems/Problem 13]]
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*[[2001 AMC 10 Problems/Problem 19]]
 
*[[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]]
*[[1986 AIME Problems/Problem 13]]
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*[[2018 AMC 8 Problems/Problem 23]]
 
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*[[2019 AMC 8 Problems/Problem 25]]
{{wikify}}{{stub}}
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*[[2018 AMC 10A Problems/Problem 11]]
 
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*[[2020 AMC 10B Problems/Problem 25]]
[[Category:Combinatorics]][[Category:Geometry]]
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[[Category:Combinatorics]]
 
 
[[latex]]
 

Latest revision as of 21:14, 21 December 2020

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 $n$ distinguishable bins? We can imagine this as finding the number of ways to drop $k$ balls into $n$ urns, or equivalently to arrange $k$ balls and $n-1$ dividers. For example, \[****||\]\[***|*|\]\[*|**|*\]\[...\] represent the ways to put $k=4$ objects in $n=3$ bins. The number of ways to do such is ${n+k-1 \choose k}$, or ${n+k-1 \choose n-1}$.


Reasoning (One of Several)

Arranging $k$ *'s and $n-1$ |'s is the same as saying there are $k+n-1$ positions: \[\underbrace{\_~~ \_~~ \_~~ \_~~ \_~~ \_~~}_{k+n-1}\] and you want to fill $k$ of them with *'s and the rest of them with |'s. Thus you are choosing $k$ positions out of $n+k-1$ total positions, resulting in a total of ${n+k-1\choose k}$ ways.

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", ..., and "repeat $k-1$-th". 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.

  • $1,2,3,4 \leftrightarrow 1,2,3,4$ (no repeats).
  • $1,1,2,4 \leftrightarrow 1,2,4,$ $r_1$.
  • $1,2,2,2 \leftrightarrow 1,2,$ $r_2$, $r_3$.
  • $4,4,5,5 \leftrightarrow 4,5,$ $r_1$, $r_2$.

Since the re-framed 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 ${n+k-1  \choose k}.$ Note: Due to the principle that ${a \choose b}={a \choose a-b}$, we can say that ${n+k-1 \choose k}={n+k-1 \choose n-1}$.

Problems

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