Stars and bars (combinatorics): Difference between revisions

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{{short description|Graphical aid for deriving some concepts in combinatorics}}

In [[combinatorics]], '''stars and bars''' (also called "sticks and stones",<ref>{{Cite book|last=Batterson|first=J|title=Competition Math for Middle School|publisher=Art of Problem Solving}}</ref> "balls and bars",<ref>{{cite book|last1=Flajolet|first1=Philippe|last2=Sedgewick|first2=Robert|date=June 26, 2009|title=Analytic Combinatorics|publisher=Cambridge University Press|isbn = 978-0-521-89806-5}}</ref> and "dots and dividers"<ref name=":0">{{Cite web|title=Art of Problem Solving|url=https://artofproblemsolving.com/wiki/index.php/Ball-and-urn|access-date=2021-10-26|website=artofproblemsolving.com}}</ref>) is a graphical aid for deriving certain [[combinatorial]] theorems. It can be used to solve manya simplevariety of [[combinatorial enumeration|counting problems]], such as how many ways there are to put {{mvar|n}} indistinguishable balls into {{mvar|k}} distinguishable bins.<ref>{{cite book |last=Feller |first=William |author-link=William Feller |year=1968 |title=An Introduction to Probability Theory and Its Applications |publisher=Wiley |volume=1 |edition=3rd |page=38}}</ref> The solution to this particular problem is given by the binomial coefficient <math>\tbinom{n+k-1}{k-1}</math>, which is the number of subsets of size {{math|''k'' − 1}} that can be formed from a set of size {{math|''n'' + ''k'' − 1}}.

If, for example, there are two balls and three bins, then the number of ways of placing the balls is <math>\tbinom{2+3-1}{3-1} = \tbinom{4}{2} = 6</math>. The table shows the six possible ways of distributing the two balls, the strings of stars and bars that represent them (with stars indicating balls and bars separating bins from one another), and the subsets that correspond to the strings. As two bars are needed to separate three bins and there are two balls, each string contains two bars and two stars. Each subset indicates which of the four symbols in the corresponding string is a bar.

{| class="wikitable"

|+ Six configurations of two balls in three bins and their star and bar represetationsrepresentations

|-

! Bin 1 !! Bin 2 !! Bin 3 !! String !! Subset of {1,2,3,4}

|-

| 2 || 0 || 0 || ★ ★ &#124;{{pipe}} &#124;{{pipe}} || {3,4}

|-

| 1 || 1 || 0 || ★ &#124;{{pipe}} ★ &#124;{{pipe}} || {2,4}

|-

| 1 || 0 || 1 || ★ &#124;{{pipe}} &#124;{{pipe}} ★ || {2,3}

|-

| 0 || 2 || 0 || &#124;{{pipe}} ★ ★ &#124;{{pipe}} || {1,4}

|-

| 0 || 1 || 1 || &#124;{{pipe}} ★ &#124;{{pipe}} ★ || {1,3}

|-

| 0 || 0 || 2 || &#124;{{pipe}} &#124;{{pipe}} ★ ★ || {1,2}

|}

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where the [[Multiset#Counting multisets|multiset coefficient]] <math>\left(\!\!\binom{k}{n}\!\!\right)</math> is the number of multisets of size {{mvar|n}}, with elements taken from a set of size {{mvar|k}}.

This corresponds to [[Composition (combinatorics)|weak compositions]] of an integer. With {{mvar|k}} fixed, the numbers for {{math|''n'' {{=}} 0, 1, 2, 3, …...}} are those in the {{math|(''k'' − 1)}}st diagonal of [[Pascal's triangle]]. For example, when {{math|''k'' {{=}} 3}} the {{mvar|n}}th number is the {{math|(''n'' + 1)}}st [[triangular number]], which falls on the second diagonal, 1, 3, 6, 10, …....

==Proofs via the method of stars and bars==

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|align=center

|content=

{{nowrap|{{huge|★ ★ ★ ★ &#124;{{pipe}} ★ &#124;{{pipe}} ★ ★}}}}

|caption=Fig.&nbsp;2: These two bars give rise to three bins containing 4, 1, and 2 objects

}}

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|align=center

|content=

{{nowrap|{{huge|★ ★ ★ ★ &#124;{{pipe}} &#124;{{pipe}} ★ &#124;{{pipe}} ★ ★ &#124;{{pipe}}}}}}

|caption=Fig.&nbsp;3: These four bars give rise to five bins containing 4, 0, 1, 2, and 0 objects

}}