Difference between revisions of "Ball-and-urn"

Latest revision as of 00:11, 5 October 2024

The ball-and-urn technique, also known as stars-and-bars, sticks-and-stones, or dots-and-dividers, 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 $4$ objects in $3$ bins. The number of ways to do such is ${4+3-1 \choose 3-1} = {6 \choose 2} = 15$ .

More generally, the number of ways to put $k$ objects into $n$ bins is ${n+k-1 \choose n-1}$ or $\dbinom{n+k-1}k$ .

The number of ways to put $k$ objects into $n$ bins, where each bin must have at least 1 object in it, is ${k-1 \choose n-1}$ .

Restrictions

Main article: Distinguishability

Let's say that we want to put $N$ objects in $n$ bins, but there must be at least $k$ objects in each bin. Without the restriction, we can set the following equation up: $a_1+a_2+a_3+\dots+a_n=N$ . We know that each $a_i$ (the bins) must have at least $k$ objects in them, so we can subtract $nk$ from $N$ , since that's how many objects are left. This means that there are $\binom{N-nk+n-1}{n-1}$ ways to distribute the objects.

Well, what if we can have at most $k$ objects in each bin? We can use the following formula to find this: $\binom{N+n-1}{n-1} - \sum_{m=1}^{\left\lfloor\frac{N}{k+1}\right\rfloor}(-1)^{m+1} \binom{n}{m} \binom{N-m(k+1)+n-1}{n-1} = \sum_{m=0}^{\left\lfloor\frac{N}{k+1}\right\rfloor}(-1)^{m} \binom{n}{m} \binom{N-m(k+1)+n-1}{n-1}$ This can be derived using the Principle of Inclusion-Exclusion.

Reasoning 1

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, or equivalently, $n-1$ positions for the dividers out of $(n-1)+k$ total spots, resulting in a total of ${n+k-1\choose n-1}$ ways.

Reasoning 2

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. $n$ 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

@@ Line 1: / Line 1: @@
-The '''ball-and-urn''' technique, also known as '''stars-and-bars''', is a commonly used technique in [[combinatorics]].
+The '''ball-and-urn''' technique, also known as '''stars-and-bars''', '''sticks-and-stones''', or '''dots-and-dividers''', 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>, given there can be <math>0</math> balls in an urn. However, if there can only be a positive amount of balls in an urn, then the number of ways to do such is <math>{k-1 \choose n-1}</math>
+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>4</math> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>.
+More generally, the number of ways to put <math>k</math> objects into <math>n</math> bins is <math>{n+k-1 \choose n-1}</math> or <math>\dbinom{n+k-1}k</math>.
+The number of ways to put <math>k</math> objects into <math>n</math> bins, where each bin must have at least 1 object in it, is <math>{k-1 \choose n-1}</math>.
+==Restrictions==
+<i> Main article: [[Distinguishability]] </i>
+Let's say that we want to put <math>N</math> objects in <math>n</math> bins, but there must be at least <math>k</math> objects in each bin. Without the restriction, we can set the following equation up: <math>a_1+a_2+a_3+\dots+a_n=N</math>. We know that each <math>a_i</math> (the bins) must have at least <math>k</math> objects in them, so we can subtract <math>nk</math> from <math>N</math>, since that's how many objects are left. This means that there are <math>\binom{N-nk+n-1}{n-1}</math> ways to distribute the objects.
+Well, what if we can have at most <math>k</math> objects in each bin? We can use the following formula to find this: <cmath>\binom{N+n-1}{n-1} - \sum_{m=1}^{\left\lfloor\frac{N}{k+1}\right\rfloor}(-1)^{m+1} \binom{n}{m} \binom{N-m(k+1)+n-1}{n-1} = \sum_{m=0}^{\left\lfloor\frac{N}{k+1}\right\rfloor}(-1)^{m} \binom{n}{m} \binom{N-m(k+1)+n-1}{n-1}</cmath> This can be derived using the [[Principle of Inclusion-Exclusion]].
+== Reasoning 1 ==
+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, or equivalently, <math>n-1</math> positions for the dividers out of <math>(n-1)+k</math> total spots, resulting in a total of <math>{n+k-1\choose n-1}</math> ways.
+== Reasoning 2 ==
+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.<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.
+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.
+*<math>1,2,3,4 \leftrightarrow 1,2,3,4</math> (no repeats).
+*<math>1,1,2,4 \leftrightarrow 1,2,4,</math> <math>r_1</math>.
+*<math>1,2,2,2 \leftrightarrow 1,2,</math> <math>r_2</math>, <math>r_3</math>.
+*<math>4,4,5,5 \leftrightarrow 4,5,</math> <math>r_1</math>, <math>r_2</math>.
+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
+\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 ==
+*[[1986 AIME Problems/Problem 13]]
+*[[2001 AMC 10 Problems/Problem 19]]
+*[[2003 AMC 10A Problems/Problem 21]]
+*[[Mock AIME 3 Pre 2005 Problems/Problem 2]]
+*[[2007 AIME I Problems/Problem 10]]
+*[[2018 AMC 8 Problems/Problem 23]]
+*[[2019 AMC 8 Problems/Problem 25]]
+*[[2018 AMC 10A Problems/Problem 11]]
+*[[2020 AMC 10B Problems/Problem 25]]
+[[Category:Combinatorics]]

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