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Combinatorics Through Guided Discovery, 2004a

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184 D. Hints to Selected Problems<br />

Additional Hint: Can you see a way to use Problem 194?<br />

197. Look for a power of a polynomial to get started.<br />

Additional Hint: The polynomial referred to in the first hint is a quotient of<br />

two polynomials. The power of the denominator can be written as a power<br />

series.<br />

198. Intepret Problem 197 in terms of multisets.<br />

199.e. When you factor out x 1 x 2 ···x n from the enumerator of trees, the result is a<br />

sum of terms of degree n − 2. (The degree of x i 1<br />

1<br />

x i 2<br />

2 ···x i n<br />

n is i 1 + i 2 + ···+ i n .)<br />

Additional Hint: Write down the picture (using xs) of a tree on five vertices<br />

with two vertices of degree one, of one with three vertices of degree one, and<br />

with four vertices of degree 1. Factor x 1 x 2 x 3 x 4 x 5 out of the picture and look<br />

at what is left. How is it related to your vertices of degree one? Now remove<br />

the vertices of degree 1 from the tree and write down the picture of the tree<br />

that remains. What is special about the vertices of degree 1 of that tree. (You<br />

can just barely learn something from this with five vertex trees, so you might<br />

want to experiment a bit with six or seven vertex trees.)<br />

200. This is a good place to apply the product principle for picture enumerators.<br />

201.a. The product principle for generating functions helps you break the generating<br />

function into a product of ten simpler ones.<br />

201.b. m was 10 in the previous part of this problem.<br />

202. Think about conjugate partitions.<br />

203.a. Don’t be afraid of writing down a product of infinitely many power series.<br />

203.b. From the fifth factor on, there is no way to choose a q i that has i nonzero and<br />

less than five from the factor.<br />

203.d. Describe to yourself how to get the coefficient of a given power of q.<br />

204. If infinitely many of the polynomials had a nonzero coefficient for q, would<br />

the product make any sense?<br />

205. (1 + q 2 + q 4 )(1 + q 3 + q 9 ) is the generating function for partitions of an integer<br />

into at most two twos and at most two threes.<br />

206. (1 + q 2 + q 4 )(1 + q 3 + q 9 ) is the generating function for partitions of an integer<br />

into at most two twos and at most two threes. (This is intentionally the same<br />

hint as in the previous problem, but it has a different point in this problem.)

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