222m2s-practice - Solution Ideas for 201001 Math 222...

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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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222m2s-practice - Solution Ideas for 201001 Math 222...

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