BULLETIN (New Series) OF THE AMERICAN MATHEMATICAL SOCIETY

The fundamental problem of enumerative combinatorics is to determine the number of elements of a set. More precisely, given an infinite indexed collection {Ai}i∈I of finite sets, we want to find a formula for the cardinality of Ai as a function of i, or at least a method for determining the number of elements of Ai that is easier (or more interesting) than counting them one at a time. There are many interesting mathematical problems that can be phrased as counting problems but to which the methods of enumerative combinatorics do not apply; for example, problems of counting groups will generally use the techniques of group theory rather than of enumerative combinatorics, and some seemingly reasonable problems, such as counting transitive relations on a set of n elements, turn out not to be amenable to the methods of enumerative combinatorics. But surprisingly many different kinds of objects can be counted, and counting them often involves interesting mathematics. The two fundamental tools of enumerative combinatorics are bijections and generating functions. Two sets A and B have the same cardinality if and only if there is a bijection from A to B, so a bijection from A to B allows us to count the elements of A by counting the elements of B. As a very simple example, a composition of an integer m is a sequence (a1, a2, . . . , ak) of positive integers with sum m. There is a bijection from the set of compositions of n + 1 to the set of subsets of {1, 2, . . . , n}: the composition (a1, . . . , ak) corresponds to the subset {a1, a1 + a2, . . . , a1 + · · · + ak−1}. So from the fact that there are 2 subsets of {1, 2 . . . , n}, we get that there are 2 compositions of n+ 1. A more interesting example is given by the enumeration of Dyck words, which are words in the letters X and Y with as many X ’s as Y ’s, and with the property that any initial segment contains at least as many X ’s as Y ’s. For example, XXYXY Y is a Dyck word, but XXYXY Y Y X is not. If we replace each X with a left parenthesis and each Y with a right parenthesis, then a Dyck word becomes a sequence of paired parentheses; thus XXYXY Y becomes (()()). We can count Dyck words of length 2n by starting with the ( 2n n ) words made up of n X ’s and n Y ’s and subtracting the number of these words that are not Dyck words. A bijection due to D. André [2] shows that ( 2n n−1 ) of these words are not Dyck words: Given a word with n X ’s and n Y ’s that is not a Dyck word, find the first Y that violates the condition and interchange all X ’s and Y ’s that occur after this Y . (For example, XXY Y Y XXY becomes XXY Y Y Y Y X .) We obtain a word with n − 1 X ’s and n + 1 Y ’s; this gives a bijection from the set of words with n X ’s and n Y ’s that are not Dyck words to the set of all words with n− 1 X ’s and n + 1 Y ’s. Thus the number of Dyck words of length 2n is the Catalan number Cn = ( 2n n ) − ( 2n n−1 ) = 1 n+1 ( 2n n ) .