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Unformatted text preview: Math 55: Discrete Mathematics Solutions for the Final Exam UC Berkeley, Spring 2009 1. (a) There are 2 nonisomorphic unrooted trees with 4 vertices: the 4 chain and the tree with one trivalent vertex and three pendant vertices. (b) There are 4 nonisomorphic rooted trees with 4 vertices, since we can pick a root in two distinct ways from each of the two trees in (a). [# 12 in § 10.1, page 694] 2. (a) There are ( 5 3 ) = 10 ways to put three balls into one box and one ball each into the other two boxes, and there are 15 = 5 · 3 ways to put one ball into one box and two balls each into the other two boxes. Hence the total number is 10 + 15 = 25 . (b) We must place one ball into each of the three boxes. Afterwards, we are left with two unlabeled balls to be placed into three labeled boxes. The number of ways to do this is the number of 2combinations from a set with 3 elements, so the answer is ( 2+3 1 2 ) = ( 4 2 ) = 6 ....
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This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.
 Spring '08
 STRAIN
 Math

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