hw5sol - Math 146: Algebraic Combinatorics Homework 5...

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Unformatted text preview: Math 146: Algebraic Combinatorics Homework 5 Solutions A hint on problem 3.1, which is quite difficult otherwise: Given a partition on n into odd parts, you can center its Ferrers diagram. Then you can divide the centered diagram into L shapes, which is then a partition of the same n into distinct parts. For example, given 18 = 7 + 5 + 3 + 3, the centered diagram is to yield 18 = 7 + 6 + 4 + 1. You should describe more rigorously what the L shapes do, and prove that it is a bijection between the two types of partitions. 3.1 Give an explicit 1- 1 correspondence between partitions of n into distinct parts and partitions of n into odd parts. Solution. Note: This proof does not use the hint provided above. Let = ( 1 ,..., k ) be a partition of n so that n = 1 + + k and each i is odd. Produce a partition from by reading left to right and each time you encounter a repeated part say i = i + 1 , add those two parts together. Repeat this process until you get (by construction) a partition of n with no repeated parts, ie a partition of n with distinct parts. To invert this procedure, let = 1 + 2 + + be a partition with distinct parts. Read left to right and each time you see an even part divide it into two equal parts of half the size. Repeat as needed until you get (by construction) a partition with only odd parts. The two maps above display a bijection between partitions of n with odd parts and partitions of n with distinct parts. 3.21 In a country that has 1-cent, 2-cent, and 3-cent coins only, show that the number of ways of changing n cents is exactly the integer nearest to ( n + 3 ) 2 12 . Solution. Let f ( n ) be the number of ways of changing n with the given coins. Then f ( n ) is the coefficient of x n in the expansion of ( 1 + x + x 2 + x 3 + )( 1 + x 2 + x 4 + x 6 + )( 1 + x 3 + x 6 + x 9 + ) . Thus we have the generating function n f ( n ) x n = n x n ! n x 2 n ! n x 3 n ! = 1 ( 1- x )( 1- x 2 )( 1- x 3 ) = 1 ( 1- x ) 3 ( 1 + x )( 1 + x + x 2 ) . Partial fractions yeilds (here can be determined using the quadratic formula on 1 + x + x 2 , and is its complex conjugate) 1 ( 1- x )( 1- x 2 )( 1- x 3 ) = 1 6 ( 1- x ) 3 + 1 4 ( 1- x ) 2 + 17 72 ( 1- x ) + 1 8 ( 1 + x ) + 1 9 ( 1- x ) + 1 9 ( 1- x ) . 1 Now if we expand each fraction on the right-hand-side as a power series and read off the coefficient of x n we get the formula f ( n ) = 1 6 n + 2 2 + 1 4 ( n + 1 )+ 17 72 + (- 1 ) n 8 + 2 9 cos ( 2 n 3 ) which can be simplified as f ( n ) = ( n + 3 ) 2 12 +- 7 + 9 (- 1 ) n + 16cos ( 2 n / 3 ) 72 where the second term is always less that 1 / 2 in absolute value. Thus f ( n ) is the unique integer whose distance from ( n + 3 ) 2 / 12 is less than 1 / 2 as required....
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hw5sol - Math 146: Algebraic Combinatorics Homework 5...

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