Difference between revisions of "Ball-and-urn"

Revision as of 01:08, 8 May 2022

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 $k=4$ 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 Two)

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 (Two of Two)

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

@@ Line 1: / Line 1: @@
 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> 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-1}</math>, or <math>{n+k-1 \choose n}</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>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>.
 == Reasoning (One of Two) ==
-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-1}</math> ways.
+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.
 == Reasoning (Two of Two) ==
@@ Line 18: / Line 18: @@
 *<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-1</math> balls that can each only go in one urn, the number of possible scenarios is simply <math>{n+k-1
+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-1}.</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-1}={n+k-1 \choose n}</math>.
+\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>.
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Difference between revisions of "Ball-and-urn"

Revision as of 01:08, 8 May 2022

Reasoning (One of Two)

Reasoning (Two of Two)

Problems