sol9 - AMS 301 (Fall 2009) Homework Set # 9 – Solution...

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Unformatted text preview: AMS 301 (Fall 2009) Homework Set # 9 – Solution Notes # 2, 6.1: (b). g (x) = (x + x2 + x3 + x4 )3 (c) (x3 + x5 )(x2 + x4 + x6 )(x2 + x3 + · · · + x6 )2 . # 4 6.1: (c). (x + x2 + x3 + . . .)7 # 14, 6.1: (a) g (x) = (x + x3 + x5 )3 (x2 + x4 + x6 )3 (b) g (x) = (x2 + x3 + x4 + x5 + x6 )(x + x3 + x4 + x5 + x6 )(x + x2 + x4 + x5 + x6 )(x + x2 + x3 + x5 + x6 )(x + x2 + x3 + x4 + x6 )(x + x2 + x3 + x4 + x5 ) # 4, 6.2: Write as x11 1 − x5 1+5 1 2+5 2 = x11 (1 − x5 ) 1 + x+ x + ··· 5 1 1 − x (1 − x) 2 2+5 7+5 + (−1) · 2 7 giving a coefficient of 1· # 20, 6.2: Let ei be the number of rooms painted color i, where i = 1, 2, 3, 4, 5. Then, we want to know how many different integer solutions there are to the system e1 + e2 + e3 + e4 + e5 = 10, subject to 0 ≤ e1 ≤ 3, 0 ≤ e2 ≤ 3, 0 ≤ e3 ≤ 3, and e4 , e5 ≥ 0. (Here, I consider colors 1, 2, and 3 to be “green”, “blue”, and “red”.) Thus, we want to know the coefficient of x10 in the generating function g (x) = (1 + x + x2 + x3 )3 (1 + x + x2 + · · ·)2 = 1 1 (1 − x4 )3 ( ) = (1 − x4 )3 ( ) 3 (1 − x)2 (1 − x) (1 − x)5 1+5−1 2+5−1 2 = (1 − 3x4 + 3x8 − x12 )(1 + x+ x + · · ·) 1 2 The coefficient of x10 is found by applying formula (6): 1· 10 + 5 − 1 6+5−1 2+5−1 + (−3) · +3· 10 6 2 ...
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This note was uploaded on 08/08/2010 for the course AMS 301 taught by Professor Arkin during the Spring '08 term at SUNY Stony Brook.

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