An Introduction to Combinatorics and Graph Theory [Lecture by David Guichard

By David Guichard

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Extra info for An Introduction to Combinatorics and Graph Theory [Lecture notes]

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Note that the first two trees have no left child, since the only tree on 0 vertices is empty, and likewise the last two have no right child. 3 • • • • • • • The 3-vertex binary rooted trees. ∞ i Now we use a generating function to find a formula for Cn . Let f = i=0 Ci x . n Now consider f 2 : the coefficient of the term xn in the expansion of f 2 is i=0 Ci Cn−i , corresponding to all possible ways to multiply terms of f to get an xn term: C0 · Cn xn + C1 x · Cn−1 xn−1 + C2 x2 · Cn−2 xn−2 + · · · + Cn xn · C0 .

Suppose the lengths of the cycles in σ are l1 , l2 , . . , lk . In cycle number i, n may be added after any of the li elements in the cycle. Thus, the total number of places that n can be added is l1 + l2 + · · · + lk = n − 1, so there are (n − 1) · n−1 permutations of [n] in which (n) is not a 1-cycle. 4 n−1 k−1 + (n − 1) · n−1 k , as desired. s(n, k) = s(n − 1, k − 1) − (n − 1)s(n − 1, k). The Stirling numbers satisfy two remarkable identities. 5 The Kronecker delta δn,k is 1 if n = k and 0 otherwise.

1 The Stirling number of the second kind, S(n, k) or number of partitions of [n] = {1, 2, . . , n} into exactly k parts, 1 ≤ k ≤ n. n k , is the Before we define the Stirling numbers of the first kind, we need to revisit permutations. 7, we may think of a permutation of [n] either as a reordering of [n] or as a bijection σ: [n] → [n]. There are different ways to write permutations when thought of as functions. Two typical and useful ways are as a table, and in cycle form. Consider this permutation σ: [5] → [5]: σ(1) = 3, σ(2) = 4, σ(3) = 5, σ(4) = 2, σ(5) = 1.

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