MATH 387 Practice Final Exam Solutions 1. (12 points) For the ...

MATH 387Practice Final Exam Solutions(b) (4 points) Construct a generating function for the number of ways to distribute n stampsin this manner. The generating function need not be algebraically simplified.Each person can get between 2 and 8 stamps, and each number of stamps can only beachieved in one way (since the stamps are all identical); thus, the generating function fordistributing stamps to a single individual is (z 2 +z 3 +z 4 +z 5 +z 6 +z 7 +z 8 ), and multiplyingtogether each individual’s stamp-selection finction, we get (z 2 +z 3 +z 4 +z 5 +z 6 +z 7 +z 8 ) 4 .(c) (4 points) Using algebraic manipulations of your generating function, determine its z 20coefficient.(z 2 +z 3 +z 4 +z 5 +z 6 +z 7 +z 8 ) 4 = (z 2 ) 4 (1+z +z 2 +z 3 +z 4 +z 5 +z 6 ) 4= z 8 ( 1−z71−z) 4= z 8 (1−z 7 ) 4 1(1−z) 4= z 8 (1−4z 7 +6z 14 −4z 21 +z 28 )= (z 8 −4z 15 +6z 22 −4z 29 +z 36 )∞∑( ) n+3z n3∞∑( n+3We see two terms in this product that could contribute to a z 20 term: (z 8 ) [( ) ] 12+33 z12and (−4z 15 ) [( ) ] 5+33 z5. Adding them together, we get the term [( (153)−483)]z 20 , so thecoefficient of z 20 is ( (153)−483)= 231.Note that this is the same as the answer to part (a), since it is counting the same thing.n=0n=03)z nPage 6 of 6 April 20, 2012