# Aspects of Combinatorics: A Wide-ranging Introduction by Victor Bryant

By Victor Bryant

Construction from fundamentals and demonstrating the relationships one of the a number of branches of combinatorics, Victor Bryant provides the consequences in an easy approach. quite a few examples and workouts together with tricks and suggestions are integrated all through and serve to guide the reader to a few of the deeper result of the topic, a lot of that are frequently excluded from introductory texts.

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N} can be extended to an n x n Latin square. Proof Assume that L is a p x n Latin rectangle with p the set Ai be given by < n, and for I < i < n let Ai = {those numbers from 1, 2,.. , n not used in the ith column of L} p L additional row: eA, I eA 2 eA. Then, as we saw in the above example, finding an additional row for L to extend it to a (p + 1) x n Latin rectangle (with entries in {1,. . , n}) is equivalent to finding distinct representatives of the family W = (Al, . , An). To show that such representatives exist, or that this family has a transversal, we use Hall's theorem (page 29) and show that any r of these sets have between them at least r elements.

IX u Yl = JXJ + IYJ-IX n Yl. The 'inclusion/exclusion principle' generalises that idea and it can be stated in terms of sets or, more conveniently for us, in terms of objects having certain properties. Example In a club there are 10 people who play tennis and 15 who play squash: 6 of them play both. How many people play at least one of the sports? Solution As in the set-theoretic result quoted above, the required number here is 10+ 15- 6= 19. 0 Example In a club there are 10 people who play tennis, 15 who play squash and 12 who play badminton.

For example the girls {1, 2, 6} know between them the boys {1', 2', 3'}. Can we find a husband for each girlfrom the boys whom they know? Solution It's easy to find husbands for the girls by trial and error, but we'll apply a process which will in fact work in general (as we shall see in the proof of the next theorem). We start to choose different boys for each of the girls in any way we like until we reach a girl for whom there is no boy left to choose. For example 1 could get engaged to 1', 2 to 2', 3 to 3', 4 to 4' and 5 to 5'.