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Unformatted text preview: Solution Ideas for 201001 Math 222 Practice MT2 Questions 1. We want the total number of sequences (10 9 ) minus the number where at least one of 1, 2, 3, 4 does not appear. For i = 1 , 2 , 3 , 4, let P i be the set of sequences in which i never appears. We need  P 1 ∪ P 2 ∪ P 3 ∪ P 4  . By counting,  P i  = 9 9 ,  P i ∩ P j  = 8 9 ,i 6 = j ,  P i ∩ P j ∩ P k  = 7 9 ,i 6 = j 6 = k , and  P 1 ∩ P 2 ∩ P 3 ∩ P 4  = 6 9 . Thus,  P 1 ∪ P 2 ∪ P 3 ∪ P 4  = ( 4 1 ) 9 9 ( 4 2 ) 8 9 + ( 4 3 ) 7 9 ( 4 4 ) 6 9 , so that the number of sequences where each of 1, 2, 3, and 4 appears is 10 9 4 1 9 9 4 2 8 9 + 4 3 7 9 4 4 6 9 = 4 X k =0 ( 1) k 4 k (10 k ) 4 . 2. We want the total number of circular permutations of 1 , 2 ,...,n (i.e. ( n 1)!) minus the number where some integer i is immediately followed by i + 1, or n is immediately followed by 1. For i = 1 , 2 ,...,n 1 let P i be the set of circular arrangements in which i is immediately followed by i + 1, and let P n be the set of circular arrangements in which n is immediately followed by 1. We need  P 1 ∪ P 2 ∪ ··· ∪ P n  . Given k < n subscripts i 1 ,i 2 ,...,i k , gluing the appropriate elements together in a block and then arranging the resulting collection of ( n k ) objects gives  P i 1 ∩ P i 2 ∩ ··· ∩ P i k  = ( n k )! / ( n k ) = ( n k 1)! (each time two numbers are glued together there is one fewer object to arrange). When k = n we have  P 1 ∩ P 2 ∩ ··· ∩ P n  = 1. Thus  P 1 ∪ P 2 ∪ ··· ∪ P n  = ∑ n 1 k =1 ( 1) k +1 ( n k...
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This note was uploaded on 01/15/2012 for the course MATH 222 taught by Professor Huangjing during the Spring '11 term at University of Victoria.
 Spring '11
 HuangJing
 Math

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