## David Guichard's An Introduction to Combinatorics and Graph Theory [Lecture PDF

By David Guichard

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The example shows that R(3) = 6. More generally, R(i, j) is the smallest integer n such that when the edges of Kn are colored with two colors, say C1 and C2 , either there is a Ki contained within Kn all of whose edges are color C1 , or there is a Kj contained within Kn all of whose edges are color C2 . Using this notion, R(k) = R(k, k). More generally still, R(i1 , i2 , . . , im ) is the smallest integer n such that when the edges of Kn are colored with m colors, C1 , . . , Cm , then for some j there is a Kij contained in Kn all of whose edges are color Cj .

128 3 3! 1) = + . 1 Newton’s Binomial Theorem not a non-negative integer, ∞ r i (x + 1)r = x i i=0 For any real number r that is when −1 < x < 1. Proof. It is not hard to see that the series is the Maclaurin series for (x + 1)r , and that the series converges when −1 < x < 1. It is rather more difficult to prove that the series is equal to (x + 1)r ; the proof may be found in many introductory real analysis books. 2 Expand the function (1 − x)−n when n is a positive integer. We first consider (x + 1)−n ; we can simplify the binomial coefficients: (−n)(−n − 1)(−n − 2) · · · (−n − i + 1) (n)(n + 1) · · · (n + i − 1) = (−1)i i!

Note that (1, 3, 5), (3, 5, 1), and (5, 1, 3) all mean the same thing. We allow 1-cycles to count as cycles, though sometimes we don’t write them explicitly. In some cases, however, it is valuable to write them to force us to remember that they are there. Consider this permutation: 13 24 35 42 51 66 . If we write this in cycle form as (1, 3, 5)(2, 4), which is correct, there is no indication that the underlying set is really [6]. Writing (1, 3, 5)(2, 4)(6) makes this clear. We say that this permutation has 3 cycles, even though one of them is a trivial 1-cycle.

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

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