C__Documents_and_Settings_raghu_Desktop_CS_2800_hw6_hw6-soln

C__Documents_and_Settings_raghu_Desktop_CS_2800_hw6_hw6-soln...

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Unformatted text preview: CS 2800 Fall 2008 Homework #6 Solution Section 5.1 Problem 32 [4pts] For each element in the domain, we have n elements to choose from in the co-domain. This choice needs to be repeated 10 times, for each element in the domain. (a) 2 10 (b) 3 10 (c) 4 10 (d) 5 10 Problem 38 [4pts] Number of subsets with more than one element = Total number of subsets - Number of subsets with one element - Number of subsets with no elements = 2 100- 100- 1 = 2 100- 101. Section 5.2 Problem 16 [5pts] We first group the numbers in the given set into four pairs that add up to 16 each, namely { 1 , 15 } , { 3 , 13 } , { 5 , 11 } , { 7 , 9 } . Now we see that if we were to pick 5 numbers from the original set, then by the pigeonhole principle, two of them must belong to the same pair, and hence we are guaranteed at least one pair of numbers that add up to 16. Problem 26 [5pts] We begin by noting that even though the result from exercise 25 applies to groups of size exactly 10, it also applies to groups of size greater than 10 because we can simply remove individuals from the group until the group size reaches 10 and then use the result. Adding back the removed individuals does not destroy the pre-existing cliques of friends/enemies. Now, pick an arbitrary person x in the group of 20 people. By definition, x is related to all the other 19 people in the group, by either a friend link or an enemy link. Since the links come in only two flavours, by the pigeon-hole principle, we can conclude that x has at least d 19 2 e = 10 links of the same type i.e. x either has at least 10 friends or 10 enemies....
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C__Documents_and_Settings_raghu_Desktop_CS_2800_hw6_hw6-soln...

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