hw12sol

# hw12sol - Math 55 Discrete Mathematics Fall 2008 Homework...

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Math 55: Discrete Mathematics, Fall 2008 Homework 12 Solutions 7.5: 8. 270 - 64 - 94 - 58 + 26 + 28 + 22 - 14 = 116 * 14. 26! - 23! - 23! - 24! + 21!. In this problem our three sets are A ﬁsh = { strings containing ﬁsh } , and A rat , A bird deﬁned similarly. We get | A ﬁsh | = 23! by treat- ing ﬁsh as a single letter, and the others likewise. We get | A ﬁsh A rat | = 21! by treating both ﬁsh and rat as single letters. The other intersections are empty, because bird has letters in common with both ﬁsh and rat . 20. 5 · 10 4 - ( 5 2 ) 10 3 + ( 5 3 ) 10 2 - ( 5 4 ) 10 + ( 5 5 ) 1. By the binomial theorem, this is equal to (1 - 10) 5 + 10 5 = 10 5 - 9 5 = 40951. 7.6: 6. The non-square-free positive integers < 100 are divisible by 2 2 , 3 2 , 5 2 or 7 2 . Let A p be the set of those divisible by p 2 . Note that A p A q = A pq since p and q are relatively prime, and likewise for products of more than two factors. Moreover, if the product is 10, then the intersection is empty, so the only non-empty intersection that contributes to the
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## This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at Berkeley.

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