hw6sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 6 (due March 9) § 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 separate into disjoint cases 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 c - b 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) 10 3 · 5 = 5000 c) Since the non-9 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 e) 26 6 f)26 6 g) 26 4 h) 26 6 + 26 6 - 26 4 , by the Inclusion-Exclusion Principle. 1
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§ 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: There must be 99
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hw6sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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