# quiz0 - Math 239 Quiz Wednesday, June 1 Instructions: No...

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Wednesday, June 1 Math 239 Quiz Instructions: No aids of any form. 1. (7 points) (a) Determine the value of the binomial coefﬁcient ( 1 3 4 ) . (b) For all non-negative integers n , determine the coefﬁcient [ x n ](1 + 2 x 3 ) - 5 . (c) If an odd integer has weight 1 and an even integer has weight 2, determine the gener- ating function for the set S of all subsets of {1,2,3}, where the weight of a subset is the sum of the weights of the integers in it. Solution: (a) - 10/243. (b) By the binomial theorem, (1 + 2 x 3 ) - 5 = X k 0 2 k ˆ - 5 k ! x 3 k Since ˆ - m k ! = ( - 1) k ˆ m - 1 + k m - 1 ! we ﬁnd that ( - 5 k ) = ( k + 4 4 ) and therefore [ x n ](1 + 2 x 3 ) - 5 = ( - 2) n /3 ˆ n 3 + 4 4 ! if 3 divides n , and is zero otherwise. (c) We have the following table of subsets and weights subset: ; {1} {2} {3} {1,2} {1,3} {2,3} {1,2,3} weight: 0 1 2 1 3 2 3 4 whence the generating function is 1 + 2 x + 2 x 2 + 2 x 3 + x 4 . 2. (6 points) Let S denote the set of all triples ( a , b , c ) of integers such that a 1, b 2, and c 3. The weight of a triple is its sum. Prove that Φ S ( x ) = x 6 (1 - x ) - 3 .

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Math 239 Quiz Wednesday, June 1 Solution: Note that S = N 1 × N 2 × N 3 . If we deﬁne the weight of an integer i to be i , then, by the Product Lemma, Φ S ( x ) = Φ N 1 ( X ) Φ N 2 ( X ) Φ N 3 ( X ) = ( x + x 2 + x 3 +··· )( x 2 + x 3 + x 4 +··· )( x 3 + x 4 + x 5 +··· ) = xx 2 x 3 (1 + x + x 2 +··· ) 3 = x 6 (1 - x ) - 3 . 3. (6 points) Suppose Φ S ( x ) = 10 x 2 + 2 x - 1 x 3 + 2 x 2 + 5 x - 1 . If Φ S ( x ) = n 0 a n x n , write down a recurrence relation and the initial conditions for a n , where n 0. Solution: ( x 3 + 2 x 2 + 5 x - 1) Φ S ( x ) = 10 x 2 + 2 x - 1. Equating the coefﬁcients of x n on both sides: n 3 - a n + 5 a n - 1 + 2 a n - 2 + a n - 3 = 0, a n = 5 a n - 1 + 2 a n - 2 + a n - 3 n = 0 - a 0 =- 1, a 0 = 1 n = 1 - a 1 + 5 a 0 = 2, a 1 = 5 a 0 - 2 = 3 n = 2 - a 2 + 5 a 1 + 2 a 0 = 10, a 2 = 5 a 1 + 2 a 0 - 10 = 7. 4. (6 points) Let a n be the number of {0,1}-strings of length n such that each even block of 0’s is followed by exactly two 1’s and each odd block of 0’s is followed by exactly three 1’s. Show that a n = [ x n ] 1 + x 1 - x 2 - 2 x 4 . Solution: Let S be the set of binary strings such that each even block of 0’s is followed by exactly two 1’s and each odd block of 0’s is followed by exactly three 1’s. We start with the block decomposition of {0,1} * , in which elements are uniquely cre- ated: {0,1} * = {1} * ({0}{0} * {1}{1} * ) * {0} * . Page 2
Math 239 Quiz Wednesday, June 1 Following each even block of 0’s, we must have exactly two 1’s, and following each odd block of 0’s, we must have exactly three 1’s. So a decomposition for S is S = {1} * ( {00}{00} * {11} {0}{00} * {111} ) * . Since this is a restriction of the block decomposition, elements of

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## This note was uploaded on 08/13/2011 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.

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quiz0 - Math 239 Quiz Wednesday, June 1 Instructions: No...

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