 MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 27, Wednesday, November 5.

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 27, Wednesday, November 5

6.3. Partitions Homework (MATH 310#9W): Read 6.4. Do 6.3: all odd numberes problems Turn in 6.3: 2,4,16,20,22 Volunteers: ____________ Problem: 16.

Partitions A partition of a group of r identical objects divides the group into a collection of unordered subsets of various sizes. Analogously, we define a partition of the interger r to be a collection of positive integers whose sum is r. Normally we write this object as a sum ans list the integers in increasing order. 5 = 1 + 1 + 1 + 1 + 1 5 = 1 + 1 + 1 + 2 5 = 1 + 2 + 2 5 = 1 + 1 + 3 5 = 2 + 3 5 = 1 + 4 5 = 5

The Generating Function The generating function for partitions can be written as the infinite product g(x) =1/[(1 – x)(1 – x 2 )... (1 – x r )...]

Example 1 Find the generating function for a r, the number of ways to express r as a sum of distinct integers. Answer: g(x) = (1+x)(1 + x 2 )... (1 + x k )...

Example 2 Find a generating function for a r, the number of ways that we can choose 2¢, 3¢, and 5¢ stamps adding to the net value of r cents. Answer: 1/[(1 – x 2 )(1 – x 3 )(1 – x 5 )]

Example 3. Show with generating functions that every positive integer can be written as a unique sum of distinct powers of 2. Answer: The generating function g * (x) = (1 + x)(1 + x 2 )(1 + x 4 ) ( 1 + x 8 )... (1 – x) g * (x) = (1 – x)(1 + x)(1 + x 2 )... = (1 – x 2 )(1 + x 2 )(1 + x 4 )... = (1 – x 4 )(1 + x 4 )(1 + x 8 )... = (1 – x 8 ) (1 + x 8 )... =... = 1 + 0x + 0x 2 +... 0x k +... = 1.

Ferrers Diagram and Conjugate Partitions Example: 15 = 7 + 3 + 2 + 2 + 1 Ferrers Diagram is shown on the left. We we transpose the diagram we obtain the conjugate partition 15 = 5 + 4 + 2 + 1 + 1 + 1+ 1

Example 4 Show that the number of partitions of an integer r as a sum of m positive integers is equal to the number of partitions of r as a sum of integers, the largest of which is equal to m.

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