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By Stanley R.

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The outdegree sequence of o is defined analogously to that of tournaments. The number of distinct outdegree sequences of orientations of G is equal to the number of spanning forests of G. 146. [*] Let G be a graph on [n]. The degree of vertex i, denoted deg(i), is the number of edges incident to i. The (ordered) degree sequence of G is the sequence (deg(1), . . , deg(n)). , no loops or multiple edges) graphs on [n] is given by f (n) = max{1, 2d(Q)−1 }, (6) Q where Q ranges over all graphs on [n] for which every connected component is either a tree or has exactly one cycle, which is of odd length.

N) ∈ Sn+1 as a product of n − 1 transpositions (the minimum possible) is nn−2 . ) For instance, the three ways to write (1, 2, 3) are (multiplying right-to-left) (1, 2)(2, 3), (2, 3)(1, 3), and (1, 3)(1, 2). Note. It is not difficult to show bijectively that the number of ways to write some n-cycle as a product of n − 1 transpositions is (n − 1)! ” However, a direct bijection between factorizations of a fixed n-cycle such as (1, 2, . . , n) and labelled trees (say) is considerably more difficult.

For instance, f (3) = 2, corresponding to s1 s2 s1 and s2 s1 s2 . Then f (n) is equal to the number of balanced tableaux (as defined in the previous problem) of shape (n − 1, n − 2, . . , 1). 51 Note. ,1) . Any bijective proof of this difficult result would be an impressive achievement. 213. [*] Let w0 = n, n − 1, n − 2, . . , 1 ∈ Sn and p = n 2 . Define Rn = {(a1 , . . , ap ) ∈ [n − 1]p : w0 = sa1 sa2 · · · sap }, where si = (i, i + 1) as in the previous problem. For example, R3 = {(1, 2, 1), (2, 1, 2)}.

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Bijective proof problems by Stanley R.


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