HW9 - HW9 Solutions Dilip Raghavan November 17, 2008 1...

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Unformatted text preview: HW9 Solutions Dilip Raghavan November 17, 2008 1 Section 7.5 Problem 8. If the first slot is 1 or 2, then there are a n- 1 possibilities for the rest. If the first slot is 0, then out of the total a n- 1 possibilities, we need to exclude the ones starting with 12. There are a n- 3 such sequences. So a n = 2 a n- 1 + a n- 1- a n- 3 = 3 a n- 1- a n- 3 . Put g ( x ) = ∑ ∞ n = a n x n . Note that a 1 = a 2 = 0 (since we must have the pattern 012), and that a 3 = 1. From these we can calculate a =- 1. Using the recurrence relation and the inital conditions, we can get g ( x ) = 1 + 3 x 1- 3 x + x 3 . Problem 13. Let a n be the number of n –digit quaternary sequences with an odd number of 1s and an odd number of 2s. Let b n be the number of n –digit quaternary sequences with an even number of 1s and an odd number of 2s, and let c n be the number of n –digit quaternary sequences with an odd number of 1s and an even number of 2s. Note that number of n –digit quaternary sequences with an even number of 1s and an even number of 2s is 4 n- a n- b n- c n . So we have the recurrence a n = 2 a n- 1 + b n- 1 + c n- 1 b n = b n- 1 + 4 n- 1...
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This note was uploaded on 09/15/2011 for the course AMAT 356 taught by Professor Hernandez/rag during the Spring '11 term at SUNY Albany.

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HW9 - HW9 Solutions Dilip Raghavan November 17, 2008 1...

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