New PDF release: An Introduction to Combinatorics and Graph Theory [Lecture

By David Guichard

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Any choice of 3 vertices defines a triangle; we wish to show that either there is a red triangle or a blue triangle. 7, with r = 3, X = 2(3 − 1) + 1 = 5), at least three of them are the same color, call it color C; call the other color D. Let the vertices at the other ends of these three edges be v1 , v2 , v3 . If any of the edges between these vertices have color C, there is a triangle of color C: if the edge connects vi to vj , the triangle is formed by v, vi , and vj . If this is not the case, then the three vertices v1 , v2 , v3 are joined by edges of color D, and form a triangle of color D.

Bl . Each set Bi , 1 ≤ i ≤ k, is contained in exactly n/2 sets in n n/2 , and all must be among A1 , . . , Ak . On average, then, each Ai , 1 ≤ i ≤ k, contains n/2 sets among n B1 , . . , Bk . But each set Ai , 1 ≤ i ≤ k, contains exactly n/2 sets in n/2 , and so each must contain exactly n/2 of the sets B1 , . . , Bk and none of the sets Bk+1 , . . , Bl . 38 Chapter 1 Fundamentals Let A1 = Aj1 = {x1 , . . , xr } and Bk+1 = {x1 , . . , xs , ys+1 , . . , yr−1 }. Let Bim = Ajm \{xs+m } and Ajm+1 = Bim ∪{ys+m }, for 1 ≤ m ≤ r −s−1.

N + i − 1)! = (−1)i i! (n − 1)! n+i−1 n+i−1 = (−1)i = (−1)i . i n−1 Thus ∞ ∞ n+i−1 −n i n+i−1 i (x + 1) = (−1) x = (−x)i . n−1 n−1 i=0 i=0 Now replacing x by −x gives −n (1 − x) ∞ = i=0 −n So (1 − x) is the generating function for 1, ∞ · 2, . . , ∞ · n} of size i. n+i−1 i x. 1 Newton’s Binomial Theorem 53 In many cases it is possible to directly construct the generating function whose coefficients solve a counting problem. 3 Find the number of solutions to x1 + x2 + x3 + x4 = 17, where 0 ≤ x1 ≤ 2, 0 ≤ x2 ≤ 5, 0 ≤ x3 ≤ 5, 2 ≤ x4 ≤ 6.

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


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