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| {{For|names of numbers in Catalan|List of numbers in various languages#Occitano-Romance}}
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| [[File:Noncrossing partitions 5.svg|thumb|The C<sub>5</sub>=42 [[noncrossing partition]]s of a 5-element set (below the other 10 of the [[Bell number|52]] [[Partition of a set|partitions]])]]
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| In [[combinatorics|combinatorial mathematics]], the '''Catalan numbers''' form a [[sequence]] of [[natural number]]s that occur in various [[Enumeration|counting problems]], often involving [[recursion|recursively]] defined objects. They are named after the [[Belgium|Belgian]] [[mathematician]] [[Eugène Charles Catalan]] (1814–1894).
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| The ''n''<sup>th</sup> Catalan number is given directly in terms of [[binomial coefficient]]s by
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| :<math>C_n = \frac{1}{n+1}{2n\choose n} = \frac{(2n)!}{(n+1)!\,n!} = \prod\limits_{k=2}^{n}\frac{n+k}{k} \qquad\mbox{ for }n\ge 0.</math>
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| The first Catalan numbers for ''n'' = 0, 1, 2, 3, … are
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| :[[1 (number)|1]], 1, [[2 (number)|2]], [[5 (number)|5]], [[14 (number)|14]], [[42 (number)|42]], [[132 (number)|132]], 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, … {{OEIS|id=A000108}}
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| == Properties ==
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| An alternative expression for ''C''<sub>''n''</sub> is
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| :<math>C_n = {2n\choose n} - {2n\choose n-1} \quad\text{ for }n\ge 0,</math>
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| which is equivalent to the expression given above because <math>\tbinom{2n}{n+1}=\tfrac{n}{n+1}\tbinom{2n}n</math>. This shows that ''C''<sub>''n''</sub> is an [[integer]], which is not immediately obvious from the first formula given. This expression forms the basis for a [[#Second proof|proof of the correctness of the formula]].
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| The Catalan numbers satisfy the [[recurrence relation]]
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| :<math>C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\sum_{i=0}^{n}C_i\,C_{n-i}\quad\text{for }n\ge 0;</math> | |
| moreover,
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| :<math>C_n= \frac 1{n+1} \sum_{i=0}^n {n \choose i}^2.</math> | |
| This is due to the fact that <math>\tbinom{2n}{n} = \sum_{i=0}^n \tbinom{n}{i}^2,</math> since choosing ''n'' numbers from a 2''n'' set of numbers can be uniquely divided into 2 parts: choosing ''i'' numbers out of the first ''n'' numbers and then choosing ''n''-''i'' numbers from the remaining ''n'' numbers.
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| They also satisfy:
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| :<math>C_0 = 1 \quad \mbox{and} \quad C_{n+1}=\frac{2(2n+1)}{n+2}C_n,</math>
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| which can be a more efficient way to calculate them.
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| Asymptotically, the Catalan numbers grow as
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| :<math>C_n \sim \frac{4^n}{n^{3/2}\sqrt{\pi}}</math>
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| in the sense that the quotient of the ''n''th Catalan number and the expression on the right [[Limit of a function|tends towards]] 1 as ''n'' → +∞. Some sources use just <math>C_n \sim \frac{4^n}{n^{3/2}}</math>.<ref>{{Cite book|title = Introduction to Algorithms|last = Cormen|first = Thomas H.|last2 = Leiserson|first2 = Charles E.|last3 = Rivest|first3 = Ronald L.|publisher = The MIT Press|isbn = 0262031418|origyear = 1990|location = Cambridge, Massachusetts|page = 304|chapter = Dynamic Programming}}</ref> (This can be proved by using [[Stirling's approximation]] for ''n''<nowiki>!</nowiki>.)
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| The only Catalan numbers ''C''<sub>''n''</sub> that are odd are those for which ''n'' = 2<sup>''k''</sup> − 1. All others are even.
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| The Catalan numbers have an integral representation
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| :<math>C_n=\int_0^4x^n\rho(x)dx</math>
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| where <math>\rho(x)=\tfrac 1{2\pi}\sqrt{\tfrac{4-x}{x}}.</math> This means that the Catalan numbers are a solution of the [[Hausdorff moment problem]] on the interval [0, 4] instead of [0, 1]. The [[orthogonal polynomials]] having the weight function <math>\rho(x)</math> on <math>[0, 4]</math> are
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| :<math>H_n(x)=\sum_{k=0}^n{n+k \choose n-k}(-x)^k.</math>
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| == Applications in combinatorics ==
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| [[File:Tamari lattice.svg|thumb|The [[associahedron]] of order 4 with C<sub>4</sub>=14 vertices]]
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| There are many counting problems in [[combinatorics]] whose solution is given by the Catalan numbers. The book ''Enumerative Combinatorics: Volume 2'' by combinatorialist [[Richard P. Stanley]] contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases ''C''<sub>3</sub> = 5 and ''C''<sub>4</sub> = 14.
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| * ''C''<sub>''n''</sub> is the number of '''Dyck words''' of length 2''n''. A Dyck word is a [[string (computer science)|string]] consisting of ''n'' X's and ''n'' Y's such that no initial segment of the string has more Y's than X's (see also [[Dyck language]]). For example, the following are the Dyck words of length 6:
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| <div class="center"><big> XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY.</big></div>
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| * Re-interpreting the symbol X as an open [[Bracket#Parentheses|parenthesis]] and Y as a close parenthesis, ''C''<sub>''n''</sub> counts the number of expressions containing ''n'' pairs of parentheses which are correctly matched:
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| <div class="center"><big> ((())) ()(()) ()()() (())() (()()) </big></div>
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| * ''C''<sub>''n''</sub> is the number of different ways ''n'' + 1 factors can be completely [[bracket|parenthesized]] (or the number of ways of [[associativity|associating]] ''n'' applications of a [[binary operator]]). For ''n'' = 3, for example, we have the following five different parenthesizations of four factors:
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| <div class="center"><big>((ab)c)d (a(bc))d (ab)(cd) a((bc)d) a(b(cd))</big></div>
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| * Successive applications of a binary operator can be represented in terms of a full [[binary tree]]. (A rooted binary tree is ''full'' if every vertex has either two children or no children.) It follows that ''C''<sub>''n''</sub> is the number of full binary [[tree (graph theory)|trees]] with ''n'' + 1 leaves:
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| [[Image:Catalan number binary tree example.png|center]]
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| * ''C''<sub>''n''</sub> is the number of non-isomorphic ordered trees with ''n''+1 vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed left-to-right order.)<ref>Stanley p.221 example (e)</ref>
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| * ''C''<sub>''n''</sub> is the number of '''monotonic paths''' along the edges of a grid with ''n'' × ''n'' square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".
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| The following diagrams show the case ''n'' = 4:
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| [[Image:Catalan number 4x4 grid example.svg|450px|center]]
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| This can be succinctly represented by listing the Catalan elements by column height:<ref>{{cite journal|last=ˇCrepinˇsek|first=Matej|coauthors=Luka Mernik|title=AN EFFICIENT REPRESENTATION FOR SOLVING CATALAN NUMBER RELATED PROBLEMS|journal=International Journal of Pure and Applied Mathematics|year=2009|volume=56|issue=4|url=http://www.ijpam.eu/contents/2009-56-4/11/11.pdf}}</ref>
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| <div style="text-align: center;">[0,0,0,0][0,0,0,1][0,0,0,2][0,0,1,1]</div>
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| <div style="text-align: center;">[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2][0,0,2,2][0,0,1,3]</div>
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| <div style="text-align: center;">[0,0,2,3][0,1,1,3] [0,1,2,2][0,1,2,3]</div>
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| * ''C''<sub>''n''</sub> is the number of different ways a [[convex polygon]] with ''n'' + 2 sides can be cut into [[triangle]]s by connecting vertices with [[straight line]]s (a form of [[Polygon triangulation]]). The following hexagons illustrate the case ''n'' = 4:
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| [[Image:Catalan-Hexagons-example.svg|400px|center]]
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| * ''C''<sub>''n''</sub> is the number of [[Stack (data structure)|stack]]-sortable [[permutation]]s of {1, ..., ''n''}. A permutation ''w'' is called [[stack-sortable permutation|stack-sortable]] if ''S''(''w'') = (1, ..., ''n''), where ''S''(''w'') is defined recursively as follows: write ''w'' = ''unv'' where ''n'' is the largest element in ''w'' and ''u'' and ''v'' are shorter sequences, and set ''S''(''w'') = ''S''(''u'')''S''(''v'')''n'', with ''S'' being the identity for one-element sequences. These are the permutations that [[Permutation pattern|avoid the pattern]] 231.
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| * ''C''<sub>''n''</sub> is the number of permutations of {1, ..., ''n''} that avoid the pattern 123 (or any of the other patterns of length 3); that is, the number of permutations with no three-term increasing subsequence. For ''n'' = 3, these permutations are 132, 213, 231, 312 and 321. For ''n'' = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
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| * ''C''<sub>''n''</sub> is the number of [[noncrossing partition]]s of the set {1, ..., ''n''}. [[A fortiori argument|''A fortiori'']], ''C''<sub>''n''</sub> never exceeds the ''n''th [[Bell number]]. ''C''<sub>''n''</sub> is also the number of noncrossing partitions of the set {1, ..., 2''n''} in which every block is of size 2. The conjunction of these two facts may be used in a proof by [[mathematical induction]] that all of the ''free'' [[cumulant]]s of degree more than 2 of the [[Wigner semicircle law]] are zero. This law is important in [[free probability]] theory and the theory of [[random matrices]].
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| * ''C''<sub>''n''</sub> is the number of ways to tile a stairstep shape of height ''n'' with ''n'' rectangles. The following figure illustrates the case ''n'' = 4:
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| [[Image:Catalan stairsteps 4.svg|400px|center]]
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| * ''C''<sub>''n''</sub> is the number of [[Young_tableau#Tableaux|standard Young tableaux]] whose diagram is a 2-by-''n'' rectangle. In other words, it is the number of ways the numbers 1, 2, ..., 2''n'' can be arranged in a 2-by-''n'' rectangle so that each row and each column is increasing. As such, the formula can be derived as a special case of the [[Young_tableau#Dimension_of_a_representation|hook-length formula]].
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| * ''C''<sub>''n''</sub> is the number of ways that the vertices of a convex 2''n''-gon can be paired so that the line segments joining paired vertices do not intersect. This is precisely the condition that guarantees that the paired edges can be identified (sewn together) to form a closed surface of genus zero (a topological 2-sphere).
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| * ''C''<sub>''n''</sub> is the number of [[semiorder]]s on ''n'' unlabeled items.<ref>{{citation
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| | last1 = Kim | first1 = K. H.
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| | last2 = Roush | first2 = F. W.
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| | mr = 538212
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| | issue = 2
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| | journal = Journal of Combinatorics, Information &System Sciences
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| | pages = 58–61
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| | title = Enumeration of isomorphism classes of semiorders
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| | volume = 3
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| | year = 1978}}.</ref>
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| == Proof of the formula ==
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| There are several ways of explaining why the formula
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| :<math>C_n = \frac{1}{n+1}{2n\choose n}</math>
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| solves the combinatorial problems listed above. The first proof below uses a [[generating function]]. The other proofs are examples of [[bijective proof]]s; they involve literally counting a collection of some kind of object to arrive at the correct formula.
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| === First proof ===
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| We first observe that all of the combinatorial problems listed above satisfy [[Johann Andreas Segner|Segner's]]<ref>A. de Segner, Enumeratio modorum, quibus figurae planae rectilineae per diagonales dividuntur in triangula. ''Novi commentarii academiae scientiarum Petropolitanae'' '''7''' (1758/59) 203–209.</ref> [[recurrence relation]]
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| :<math>C_0 = 1 \quad \text{and} \quad C_{n+1}=\sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0.</math>
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| For example, every Dyck word ''w'' of length ≥ 2 can be written in a unique way in the form
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| :''w'' = X''w''<sub>1</sub>Y''w''<sub>2</sub>
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| with (possibly empty) Dyck words ''w''<sub>1</sub> and ''w''<sub>2</sub>.
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| The [[generating function]] for the Catalan numbers is defined by
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| :<math>c(x)=\sum_{n=0}^\infty C_n x^n.</math>
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| The two recurrence relations together can then be summarized in generating function form by the relation
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| :<math>c(x)=1+xc(x)^2;\,</math>
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| in other words, this equation follows from the recurrence relations by expanding both sides into power series. On the one hand, the recurrence relations uniquely determine the Catalan numbers; on the other hand, the generating function solution
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| :<math>c(x) = \frac{1-\sqrt{1-4x}}{2x}=\frac{2}{1+\sqrt{1-4x}}</math>
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| has a power series at 0 and its coefficients must therefore be the Catalan numbers. The chosen solution satisfies the following condition.
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| :<math>\lim_{x \to 0^+} c(x) = C_0 = 1</math>
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| The other solution has a pole at 0 and this reasoning doesn't apply to it.
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| The square root term can be expanded as a power series using the identity
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| :<math>\sqrt{1+y} = \sum_{n=0}^\infty {\frac12 \choose n} y^n = 1 - 2\sum_{n=1}^\infty {2n-2 \choose n-1} \left(\frac{-1}{4}\right)^n \frac{y^n}{n}.</math>
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| This is a special case of [[binomial theorem#Newton's generalized binomial theorem|Newton's generalized binomial theorem]]; as with the general theorem, it can be proved by computing derivatives to produce its Taylor series. Setting ''y'' = −4''x'' and substituting this power series into the expression for ''c(x)'' and shifting the summation index ''n'' by 1, the expansion simplifies to
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| :<math>c(x) = \sum_{n=0}^\infty {2n \choose n} \frac{x^n}{n+1}.</math>
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| The coefficients are now the desired formula for ''C''<sub>''n''</sub>.
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| Another way to get ''c''(''x'') is to solve for ''xc''(''x'') and observe that <math>\int_0^x \! t^n \, dt</math> appears in each term of the power series.
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| === Second proof ===
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| This proof depends on a trick known as [[André's reflection method]] (not to be confused with the [[Schwarz reflection principle]] in [[complex analysis]]), which was originally used in connection with [[Bertrand's ballot theorem]]. The reflection principle has been widely attributed to [[Désiré André]], but his method did not actually use reflections; and the reflection method is a variation due to Aebly and Mirimanoff.<ref>[http://webspace.ship.edu/msrenault/ballotproblem/monthly358-363-renault.pdf Renault, Marc, Lost (and found) in translation: André's actual method and its application to the generalized ballot problem. Amer. Math. Monthly 115 (2008), no. 4, 358--363.]</ref> It is most easily expressed in terms of the "monotonic paths which do not cross the diagonal" problem (see [[#Applications in combinatorics|above]]).
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| [[Image:Catalan number reflection example.png|frame|right|Figure 1. The green portion of the path is flipped. The old grid is ''n'' x ''n''. The new grid is (''n'' - 1) x (''n'' + 1).]]
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| Suppose we are given a monotonic path in an ''n'' × ''n'' grid that ''does'' cross the diagonal. Find the first edge in the path that lies above the diagonal, and ''flip'' the portion of the path occurring after that edge, along a line parallel to the diagonal. Observe that now we have taken into account ''k''+1 vertical edges and ''k'' horizontal edges for some ''k'' between 1 and ''n''-1. This leaves ''l''-1 vertical edges and ''l'' horizontal edges, where ''l'' + ''k'' = ''n''. (In terms of Dyck words, we are starting with a sequence of ''n'' X's and ''n'' Y's which is ''not'' a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition. At that first Y, there are ''k'' + 1 Y's and ''k'' X's for some ''k'' between 1 and ''n'' - 1.). By reflecting the rest of the vertical and horizontal edges, we will have ''k'' + 1 + ''l'' = ''n'' +1 vertical edges and ''k'' + ''l'' - 1 = ''n'' - 1 horizontal edges. The resulting path is a monotonic path in an (''n'' − 1) × (''n'' + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped.
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| Since every monotonic path in the (''n'' − 1) × (''n'' + 1) grid is of length 2''n'' and every such path involves ''n'' + 1 vertical edges in any order, the number of these paths is equal to
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| :<math>{2n\choose n+1}.</math>
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| Therefore, to calculate the number of monotonic ''n'' × ''n'' paths which do ''not'' cross the diagonal, we need to subtract this from the ''total'' number of monotonic ''n'' × ''n'' paths, so we finally obtain
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| :<math>{2n\choose n}-{2n\choose n+1}</math>
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| which is the ''n''th Catalan number ''C''<sub>''n''</sub>.
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| === Third proof ===
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| The following bijective proof, while being more involved than the previous one, provides a more natural explanation for the term ''n'' + 1 appearing in the denominator of the formula for ''C''<sub>''n''</sub>. A generalized version of this proof can be found in a paper of Rukavicka Josef (2011).<ref>Rukavicka Josef (2011), ''On Generalized Dyck Paths, Electronic Journal of Combinatorics [http://www.combinatorics.org/ojs/index.php/eljc/article/view/v18i1p40/pdf online]</ref>
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| [[Image:Catalan number exceedance example.png|frame|right|Figure 2. A path with exceedance 5.]]
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| Suppose we are given a monotonic path, which may happen to cross the diagonal. The '''exceedance''' of the path is defined to be the number of vertical edges which lie ''above'' the diagonal. For example, in Figure 2, the edges lying above the diagonal are marked in red, so the exceedance of the path is 5.
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| Now, if we are given a monotonic path whose exceedance is not zero, then we may apply the following algorithm to construct a new path whose exceedance is one less than the one we started with.
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| * Starting from the bottom left, follow the path until it first travels above the diagonal.
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| * Continue to follow the path until it ''touches'' the diagonal again. Denote by ''X'' the first such edge that is reached.
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| * Swap the portion of the path occurring before ''X'' with the portion occurring after ''X''.
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| The following example should make this clearer. In Figure 3, the black dot indicates the point where the path first crosses the diagonal. The black edge is ''X'', and we swap the red portion with the green portion to make a new path, shown in the second diagram.
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| [[Image:Catalan number swapping example.png|frame|center|Figure 3. The green and red portions are being exchanged.]]
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| Notice that the exceedance has dropped from three to two. In fact, the algorithm will cause the exceedance to decrease by one, for any path that we feed it, because the first vertical step starting on the diagonal (at the point marked with a black dot) is the unique vertical edge that under the operation passes from above the diagonal to below it; all other vertical edges stay on the same side of the diagonal.
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| [[Image:Catalan number algorithm table.png|frame|right|Figure 4. All monotonic paths in a 3×3 grid, illustrating the exceedance-decreasing algorithm.]]
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| It is also not difficult to see that this process is ''reversible'': given any path ''P'' whose exceedance is less than ''n'', there is exactly one path which yields ''P'' when the algorithm is applied to it. Indeed, the (black) edge ''X'', which originally was the first horizontal step ending on the diagonal, has become the ''last'' horizontal step ''starting'' on the diagonal.
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| This implies that the number of paths of exceedance ''n'' is equal to the number of paths of exceedance ''n'' − 1, which is equal to the number of paths of exceedance ''n'' − 2, and so on, down to zero. In other words, we have split up the set of ''all'' monotonic paths into ''n'' + 1 equally sized classes, corresponding to the possible exceedances between 0 and ''n''. Since there are
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| :<math>{2n\choose n}</math>
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| monotonic paths, we obtain the desired formula
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| :<math>C_n = \frac{1}{n+1}{2n\choose n}.</math>
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| Figure 4 illustrates the situation for ''n'' = 3. Each of the 20 possible monotonic paths appears somewhere in the table. The first column shows all paths of exceedance three, which lie entirely above the diagonal. The columns to the right show the result of successive applications of the algorithm, with the exceedance decreasing one unit at a time. There are five rows, that is, ''C''<sub>3</sub> = 5.
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| === Fourth proof ===
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| This proof uses the triangulation definition of Catalan numbers to establish a relation between ''C<sub>n</sub>'' and ''C''<sub>''n''+1</sub>. Given a polygon ''P'' with ''n''+ 2 sides, first mark one of its sides as the base. If ''P'' is then triangulated, we can further choose and orient one of its 2''n''+1 edges. There are (4''n''+2)''C<sub>n</sub>'' such decorated triangulations. Now given a polygon ''Q'' with ''n''+3 sides, again mark one of its sides as the base. If ''Q'' is triangulated, we can further mark one of the sides other than the base side. There are (''n''+2)''C''<sub>''n''+1</sub> such decorated triangulations. Then there is a simple bijection between these two kinds of decorated triangulations: We can either collapse the triangle in ''Q'' whose side is marked, or in reverse expand the oriented edge in ''P'' to a triangle and mark its new side. Thus
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| :<math>(4n+2)C_n = (n+2)C_{n+1}.</math>
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| The binomial formula for ''C<sub>n</sub>'' follows immediately from this relation and the initial condition ''C''<sub>1</sub> = 1.
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| === Fifth proof ===
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| This proof is based on the [[Dyck language|Dyck words]] interpretation of the Catalan numbers, so ''C''<sub>''n''</sub> is the number of ways to correctly match ''n'' pairs of brackets. We denote a (possibly empty) ''correct'' string with ''c'' and its inverse (where "[" and "]" are exchanged) with ''c''<sup>+</sup>. Since any ''c'' can be uniquely decomposed into ''c'' = [ ''c''<sub>''1''</sub> ] ''c''<sub>''2''</sub>, summing over the possible spots to place the closing bracket immediately gives the recursive definition
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| :<math>C_0 = 1 \quad \text{and} \quad C_{n+1} = \sum_{i=0}^n C_i\,C_{n-i}\quad\text{for }n\ge 0.</math>
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| Now let ''b'' stand for a ''balanced'' string of length 2''n''—that is, containing an equal number of "[" and "]"—and <math>\textstyle B_n = {2n\choose n} = d_n C_n</math> with some factor ''d''<sub>''n''</sub> ≥ 1. As above, any balanced string can be uniquely decomposed into either [ ''c'' ] ''b'' or ] ''c''<sup>+</sup>[''b'', so
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| :<math>B_{n+1} = 2 \sum_{i=0}^n B_i C_{n-i}.</math>
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| Also, any incorrect balanced string starts with ''c'' '']'', so
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| :<math>B_{n+1} - C_{n+1} = \sum_{i=0}^n {2i+1 \choose i} C_{n-i} = \sum_{i=0}^n \frac{2i+1}{i+1} B_i C_{n-i}.</math>
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| Subtracting the above equations and using ''B''<sub>''i''</sub> = ''d''<sub>''i''</sub> ''C''<sub>''i''</sub> gives
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| :<math>C_{n+1} = 2 \sum_{i=0}^n d_i C_i C_{n-i} - \sum_{i=0}^n \frac{2i+1}{i+1} d_i C_i C_{n-i} = \sum_{i=0}^n \frac{d_i}{i+1} C_i C_{n-i}.</math>
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| Comparing coefficients with the original recursion formula for ''C''<sub>''n''</sub> gives ''d''<sub>''i''</sub> = ''i'' + 1, so
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| :<math>C_n = \frac{1}{n+1}{2n\choose n}.</math>
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| == Hankel matrix ==
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| The ''n''×''n'' [[Hankel matrix]] whose (''i'', ''j'') entry is the Catalan number ''C''<sub>''i''+''j''−2</sub> has [[determinant]] 1, regardless of the value of ''n''. For example, for ''n'' = 4 we have
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| :<math>\det\begin{bmatrix}1 & 1 & 2 & 5 \\ 1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132\end{bmatrix} = 1.</math>
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| Moreover, if the indexing is "shifted" so that the (''i'', ''j'') entry is filled with the Catalan number ''C''<sub>''i''+''j''−1</sub> then the determinant is still 1, regardless of the value of ''n''.
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| For example, for ''n'' = 4 we have
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| :<math>\det\begin{bmatrix}1 & 2 & 5 & 14 \\ 2 & 5 & 14 & 42 \\ 5 & 14 & 42 & 132 \\ 14 & 42 & 132 & 429 \end{bmatrix} = 1.</math>
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| Taken together, these two conditions uniquely define the Catalan numbers.
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| == History ==
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| The Catalan sequence was first described in the 18th century by [[Leonhard Euler]], who was interested in the number of different ways of dividing a polygon into triangles. The sequence is named after [[Eugène Charles Catalan]], who discovered the connection to parenthesized expressions during his exploration of the [[Towers of Hanoi]] puzzle. The counting trick for Dyck words was found by D. André in 1887.
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| In 1988, it came to light in an [[Inner Mongolia University of Technology]] publication that the Catalan number sequence had been used in [[China]] by the Mongolian mathematician [[Minggantu]] by 1730. That is when he started to write his book ''Ge Yuan Mi Lu Jie Fa'', which was completed by his student Chen Jixin in 1774 but published sixty years later. P.J. Larcombe (1999) sketched some of the features of the work of Minggantu, including the stimulus of Pierre Jartoux, who brought three infinite series to China early in the 1700s.
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| For instance, Ming used the Catalan sequence to express series expansions of sin(2α) and sin(4α) in terms of sin(α).
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| == See also ==
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| {{Col-begin}}
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| {{Col-1-of-3}}
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| * [[Associahedron]]
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| * [[Bertrand's ballot theorem]]
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| * [[Binomial transform]]
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| * [[File:Wikiversity-logo-en.svg|20px]] [[v:Partition related number triangles|Partition related number triangles]]
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| {{Col-2-of-3}}
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| * [[Catalan's problem]]
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| * [[Catalan's triangle]]
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| * [[Catalan–Mersenne number]]
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| * [[Fuss–Catalan number]]
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| * [[List of factorial and binomial topics]]
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| {{Col-3-of-3}}
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| {{Portal|Mathematics}}
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| * [[Lobb numbers]]
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| * [[Narayana number]]
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| * [[Schröder–Hipparchus number]]
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| * [[Tamari lattice]]
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| {{Col-end}}
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| == Notes ==
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| {{Reflist}}
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| ==References==
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| * [[John H. Conway|Conway]] and [[Richard Guy|Guy]] (1996) ''The Book of Numbers''. New York: Copernicus, pp. 96–106.
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| * {{citation
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| | last = Gardner
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| | first = Martin
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| | authorlink = Martin Gardner
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| | title = Time Travel and Other Mathematical Bewilderments
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| | publisher = W.H. Freeman and Company
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| | year = 1988
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| | location = New York
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| | pages = 253–266 (Ch. 20)
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| | isbn = 0-7167-1924-X}}
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| * {{citation
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| | last = Koshy
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| | first = Thomas
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| | title = Catalan Numbers with Applications
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| | publisher = Oxford University Press
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| | year = 2008
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| | pages =
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| | url = http://www.amazon.com/Thomas-Koshy/e/B001H6NZT4/ref=ntt_athr_dp_pel_1
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| | isbn = 0-19-533454-X }}
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| * Koshy, Thomas & Zhenguang Gao (2011) "Some divisibility properties of Catalan numbers", [[Mathematical Gazette]] 95:96–102.
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| * Larcombe, P.J. (1999) "[http://www.math.ucla.edu/~pak/lectures/Cat/Larcombe-The_18th_century_Chinese_discovery_of_the_Catalan_numbers.pdf The 18th century Chinese discovery of the Catalan numbers]", ''[[Mathematical Spectrum]]'' 32:5–7.
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| * {{Citation | last1=Stanley | first1=Richard P. | title=Enumerative combinatorics. Vol. 2 | url=http://www-math.mit.edu/~rstan/ec/ | publisher=[[Cambridge University Press]] | series=Cambridge Studies in Advanced Mathematics | isbn=978-0-521-56069-6 | mr=1676282 | year=1999 | volume=62}}
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| * {{citation
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| | last = Egecioglu
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| | first = Omer
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| | title = A Catalan-Hankel Determinant Evaluation
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| | year = 2009
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| | pages =
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| | url = http://www.cs.ucsb.edu/~omer/DOWNLOADABLE/catalan_hankel09.pdf}}
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| ==External links==
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| * {{citation|last=Stanley|first= Richard P.|url=http://www-math.mit.edu/~rstan/ec/catadd.pdf |format=PDF|title=Catalan addendum to Enumerative Combinatorics, Volume 2|year=1998}}
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| * {{mathworld|title=Catalan Number|urlname=CatalanNumber}}
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| * Dickau, Robert M.: [http://mathforum.org/advanced/robertd/catalan.html Catalan numbers] Further examples.
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| * Davis, Tom: [http://mathcircle.berkeley.edu/BMC6/pdf0607/catalan.pdf Catalan numbers]. Still more examples.
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| * Schmidthammer, Jürgen: [http://www.bnv-bamberg.de/home/ba2636/catalanz.pdf Catalan-Zahlen] Zulassungsarbeit zum Staatsexamen (PDF-File; 7,05 MB)
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| * "Equivalence of Three Catalan Number Interpretations" from The Wolfram Demonstrations Project [http://demonstrations.wolfram.com/EquivalenceOfThreeCatalanNumberInterpretations/]
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| {{Classes of natural numbers}}
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| {{DEFAULTSORT:Catalan Number}}
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| [[Category:Integer sequences]]
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| [[Category:Factorial and binomial topics]]
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| [[Category:Enumerative combinatorics]]
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| [[Category:Articles containing proofs]]
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