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Unformatted text preview: Math 55: Discrete Mathematics UC Berkeley, Spring 2010 Solutions to Homework # 6 5.1, #8: There are 26 choices for the first letter, then 25 for the second and 24 for the third to avoid repetition. Thus there are 26 25 24 = 15600 possibilities. 5.1, #16: We want to count how many words of length four have the letter x in them. We separate into disjoint cases (according to the placement of the first x appearing in the word) and apply the Sum Rule. Either the first letter is x (leaving the others unconstrained) or the first is not x but the second is x , or the first two are not but the third is, or the first three are not but the fourth is. Thus there are 26 3 +25 26 2 +25 2 26+25 3 = 66351 different such strings. (Note: x is the first letter and x is the second letter are not disjoint cases!) 5.1, #19: a) Should be b 100 7 cb 49 7 c = 7 multiples of 7 between 50 and 100; these are { 56 , 63 , 70 , 77 , 84 , 91 , 98 } . b) b 100 11 c  b 49 11 c = 5; { 55 , 66 , 77 , 88 , 99 } . c) The numbers that are both multiples of 7 and 11 are precisely the multiples of 77; clearly 77 is the only one of these between 50 and 100. 5.1, #24: a) 10 9 8 7 = 5040 b) 5 10 3 = 5000 c) Since the non9 digit can be in one of four places (and these cases are disjoint) and can be any of the other 9 digits, there are 4 9 = 36 such strings. 5.1, #30: a) 26 8 b) 26 25 24 23 22 21 20 19 c) 26 7 d) 25 24 23 22 21 20 19 1 e) 26 6 f)26 6 g) 26 4 h) 26 6 + 26 6 26 4 , by the InclusionExclusion Principle. 5.1, #55: You can draw a tree graph in which each branching asks whether a given element belongs to the subset, each node at a given level is labeled with the sum of its elements thus far, and only nodes whose sum of elements is 28 are included. The graph (too unwieldy to picture here) indeed has 17 nodes at the bottom, corresponding to the 17 such subsets. 5.2, #9: We solve this exercise using the Pigeonhole Principle. Think of the states of the country as 50 boxes and represent the students by balls. We want to find the minimum n s.t. d fracn 50 e 100. This number is 4951. Thus, there must be 99 50 + 1 = 4951 students. We could admit only 99 from each state and have 4950 students total, but if we have 4951, then some state must contain 100 of the students....
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This note was uploaded on 04/05/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.
 Spring '08
 STRAIN
 Math

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