Introductory Combinatorics (5th Edition) by Richard A. Brualdi

By Richard A. Brualdi

This relied on best-seller emphasizes combinatorial ideas–including the pigeon-hole precept, counting concepts, diversifications and mixtures, Pólya counting, binomial coefficients, inclusion-exclusion precept, producing features and recurrence family, combinatortial buildings (matchings, designs, graphs), and flows in networks. The 5th variation clarifies the exposition all through and provides a wealth of recent routines. Appropriate for one- or two-semester, junior- to senior-level combinatorics courses.

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We now equate the two counts and complete the proof. 4 is an instance of obtaining an identity by counting the objects of a set (in this case the subsets of a set of n elements) in two different ways and setting the results equal to one another. This technique of "double counting" is a. powerful one in combinatorics, and we will see several other applications of it. Example. The number of 2-subsets of the set {I, 2, ... , n} of the first n positive integers is (~). Partition the 2-subsets according to the largest integer they contain.

This way of looking at circular permutations is also useful when the formula for circular permutations cannot be applied directly. Example. Ten people, including two who do not wish to sit next to one another, are to be seated at a round table. How many circular seating arrangements are there? We solve this problem using the subtraction principle. Let the 10 people be P l , P2 , P3 , • .. ,PlO , where P l and P2 are the two who do not wish to sit together. Consider seating arrangements for 9 people X, P3, ...

Twenty-five points are chosen in the plane so that no three of them are collinear. How many straight lines do they determine? How many triangles do they determine? Since no three of the points lie on a line, every pair of points determines a unique straight line. Thus, the number of straight lines determined equals the number of 2-subsets of a 25-element set, and this is given by 25! 23! = 300. Similarly, every three points determines a unique triangle, so that the number of triangles determined is given by 25!

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