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Unformatted text preview: MATH 239 ASSIGNMENT 2 Sample Solutions 1. Suppose Adam has a total of 19 pennies, 39 dimes, and 9 “toonies” ($ 2 coins) on his dresser. Let the weight of a coin be its monetary value. Let b n be the number of ways one can obtain a total of n cents from this collection of coins (assuming that coins of the same denomination are indistinguishable). (a) Let B ( x ) = b + b 1 x + b 2 x 2 + ... . Use the product lemma to prove that B ( x ) = (1 + x 10 )(1 + x 200 )(1 x 2000 ) 1 x . Solution: The possible selections of coins correspond to 3tuples (or ordered triples) of numbers ( p,d,t ) where p ∈ P = { , 1 , 2 , 3 ,..., 19 } , d ∈ D = { , 1 ,..., 39 } , and t ∈ T = { , 1 ,..., 9 } . In particular, p denotes the number of pennies chosen, d the number of dimes, and t the number of toonies. In other words, ( p,d,t ) ∈ P × D × T . We define the weight (or value) of a particular arrangement ( p,d,t ) of coins to be p + 10 d + 200 t , so that b n equals the number of 3tuples in P × D × T with weight ω ( p,d,t ) = n . Therefore b n is the coefficient of x n in Φ P × D × T ( x ). In other words, B ( x ) = Φ P × D × T ( x ). The value of p pennies is p , thus we use weight function ω P ( p ) = p . The generating function for P is thus Φ P ( x ) = 1 + x + ... + x 19 = 1 x 20 1 x . The value of d dimes is 10 d , thus we use weight function ω D ( d ) = 10 d . The generating function for D is thus Φ D ( x ) = 1 + x 10 + x 20 + ... + x 390 = 1 x 400 1 x 10 ....
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This note was uploaded on 08/09/2009 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
 Spring '09
 M.PEI
 Math, Combinatorics

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