MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 26, Monday, November 3.

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 26, Monday, November 3

6.2. Calculating Coefficients of Generating Functions Homework (MATH 310#9M): Read 6.3. Do 6.2: all odd numberes problems Turn in 6.2: 2,4,6,14,20,32,38 Volunteers: ____________ Problem: 38.

Polynomial Expansions (1 – x n+1 )/(1 - x) = 1 + x x n. 1/(1 – x) = 1 + x + x x n +... (1 + x) n = 1 + C(n,1)x + C(n,2)x C(n,r)x r C(n,n)x n. (1 – x m ) n = 1 - C(n,1)x m + C(n,2)x 2m (-1) r C(n,r)x rm (- 1) n C(n,n)x nm. 1/(1 – x) n = 1 + C(1+n-1,1)x + C(2 + n –1,2)x C(r + n – 1,r)x r +... If h(x) = f(x)g(x), where f(x) = a 0 + a 1 x + a 2 x and g(x) = b 0 + b 1 x + b 2 x , then h(x) = a 0 b 0 + (a 1 b 0 + a 0 b 1 )x + (a 2 b 0 + a 1 b 1 + a 0 b 2 )x

Example 1. Find the coefficient of x 16 in (x 2 + x 3 + x ) 5. Determine the coefficient of x r. Answer: a 16 = C(6 + 5 – 1,6) = C(10,4). a r = C(r ,r-10) = C(r-6,4).

Example 2 Use generating functions to find the number of ways to collect $15 from 20 distinct people if each of the first 19 can give a dollar (or nothing) and the 20th can give $1 or $5 (or nothing). Answer: (1 + x) 19 (1 + x + x 5 ), a 15 = C(19,15) £ 1 + C(19,14) £ 1 + C(19,10) £ 1 = C(19,4) + C(19,5) + C(19,9).

Example 3 How many ways are there to distribute 25 identical balls into seven distinct boxes if the first box can have no more than 10 balls but any number can go into each of the other six boxes? Answer: (1 – x 11 )(1/(1-x) 7 ). a 0 b 25 + a 11 b 14 = 1 £ C( – 1,25) + (-1) £ C( – 1, 14) = C(31,6) – C(20,6).

Example 4 How many ways are there to select 25 toys from 7 types of toys with between 2 and 6 of each type? Answer: (x 2 + x x 6 ) 7 = x 14 (1 + x + x 2 + x 3 + x 4 ) 7 = x 14 (1 –x 5 ) 7 /(1-x) 7. c 11 = a 0 b 11 + a 5 b 6 + a 10 b 1 = 1 £ C( – 1,11) + (-C(7,1)) £ C(6 + 7 – 1,6) + C(7,2) £ C(1 + 7 – 1,1) = C(17,6) – 7C(12,6) + C(7,2)7.

Example 5 – Binomial Identity Verify the binomial identity: C(n,0) 2 + C(n,1) C(n,n) 2 = C(2n,n). Answer: (1 + x) n (1 + x) n = (1 + x) 2n. c n = C(2n,n) = a 0 b n + a 1 b n a n-1 b 1 + a n b 0 = C(n,0)C(n,n) + C(n,1)C(n,n-1) C(n,n-1)C(n,1) + C(n,n)C(n,0). Since C(n,r) = C(n,n-r) the result follows.