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Chapter 3: Problem 176

Show that the number of partitions of \(k\) into distinct parts equals the number of partitions of \(k\) into odd parts. (h)

Short Answer

Expert verified
The number of partitions of k into distinct parts equals the number of partitions of k into odd parts due to a bijection between their respective sets.

Step by step solution

01

Understand the Definition of Partitions

A partition of a number is a way of writing it as a sum of other numbers. In your exercise, two types of partitions are considered: those with distinct parts (no repeated numbers) and those with only odd parts.
02

Represent Distinct Partitions

Consider the partitions of given number k into distinct parts. For example, if k=7, a distinct partition could be (4, 3) or (5, 2).
03

Convert Distinct Parts to Odd Parts

Use a known bijection: Replace each part in the partition with distinct parts by pairs of odd numbers, preserving their sum. For example, a part '4' can be replaced by '3+1'. Essentially, partition each number into 1+3, 1+1+1+1 (all odd parts).
04

Verify the Conversion

Check that each distinct part conversion results in a partition with only odd parts. For example, partition (4,3) converts to (3+1,3) which is (3, 3, 1). All parts are thus odd.
05

Count the Partitions

Confirm that each partition of k into distinct parts corresponds to a unique partition into odd parts and vice versa. This establishes a one-to-one correspondence or bijection, proving that the number of such partitions is the same in both cases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distinct Parts Partitions
In combinatorics, when we talk about 'distinct parts partitions' of a number, we refer to breaking the number down into a sum of smaller numbers that are all different from each other.
For example, if we take the number 7, its distinct partitions are (4 + 3), (5 + 2), or (6 + 1).
Notice that no number repeats in these partitions. This concept is crucial in problems where repetition is not allowed.
Understanding this helps in comparing it to other types of partitions and in finding relationships between different partition types.
When performing these partitions, every unique combination counts towards the total number of partitions.
This is the key to understanding why different partition sets can be linked or bijective.
Odd Parts Partitions
The 'odd parts partitions' of a number focus on breaking the number into parts that are all odd numbers.
For example, for the number 7, the partitions into odd parts are (7), (5 + 1 + 1), (3 + 3 + 1), or (1 + 1 + 1 + 1 + 1 + 1 + 1).
All these parts are odd numbers.
Among its interesting properties is that odd parts partitioning often reveals symmetrical and structured forms.
The restriction to odd numbers makes these partitions unique and often simpler in structure.
This forms a basis for more advanced combinatorial arguments and proofs.
It is useful in various advanced mathematical areas, including number theory and algebraic combinatorics.
Bijection in Combinatorics
A bijection in combinatorics is a one-to-one correspondence between the elements of two sets.
This means that every element in one set is paired with exactly one element in the other set, and vice versa.
In this exercise, the bijection is between the partitions of a number into distinct parts and the partitions of the same number into odd parts.
To use bijection here, convert each distinct part in a partition into an equivalent combination of odd parts.
For example, a part '4' can be written as '3 + 1', both of which are odd.
This way, each distinct partition can be mapped to a unique odd partition, proving the one-to-one relationship.
Understanding bijection helps not only in partition problems but in a wide range of combinatorial proofs and applications, providing a powerful tool for demonstrating equality of sets in terms of their cardinality.
Overall, the concept is an elegant way to show that two seemingly different types of partitions are, in fact, equivalent in count.

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Most popular questions from this chapter

Show that every power of \(x+1\) is expressible as a sum of numerical multiples of powers of \(x .\) Now show that every power of \(x\) (and thus every polynomial in \(x\) ) is a sum of numerical multiples (some of which could be negative) of powers of \(x+1\). This means that the powers of \(x+1\) are a basis for the space of polynomials as well. Describe the change of basis coefficients that we use to express the binomial powers \((x+1)^{n}\) in terms of the ordinary \(x^{j}\) explicitly. Find the change of basis coefficients we use to express the ordinary powers \(x^{n}\) in terms of the binomial powers \((x+1)^{k} .\) (h)

Explain the relationship between the number of partitions of \(k\) into even parts and the number of partitions of \(k\) into parts of even multiplicity, i.e. parts which are each used an even number of times as in (3,3,3,3,2,2,1,1)

Explain the relationship between partitions of \(k\) into \(n\) parts and lists \(x_{1}, x_{2}, \ldots, x_{n}\) of positive integers that add to \(k\) with \(x_{1} \geq x_{2} \geq\) \(\ldots \geq x_{n} .\) Such a representation of a partition is called a decreasing list representation of the partition.

Suppose we wish to place \(k\) distinct books onto the shelves of a bookcase with \(n\) shelves. For simplicity, assume for now that all of the books would fit on any of the shelves. Also, let's imagine pushing the books on a shelf as far to the left as we can, so that we are only thinking about how the books sit relative to each other, not about the exact places where we put the books. Since the books are distinct, we can think of a the first book, the second book and so on. (a) How many places are there where we can place the first book? (b) When we place the second book, if we decide to place it on the shelf that already has a book, does it matter if we place it to the left or right of the book that is already there? (c) How many places are there where we can place the second book? (h) (d) Once we have \(i-1\) books placed, if we want to place book \(i\) on a shelf that already has some books, is sliding it in to the left of all the books already there different from placing it to the right of all the books already or between two books already there? (e) In how many ways may we place the \(i\) th book into the bookcase? (h) (f) In how many ways may we place all the books?

Given a function \(f\) from a \(k\) -element set \(K\) to an \(n\) -element set, we can define a partition of \(K\) by putting \(x\) and \(y\) in the same block of the partition if and only if \(f(x)=f(y)\). How many blocks does the partition have if \(f\) is onto? How is the number of functions from a \(k\) -element set onto an \(n\) -element set related to a Stirling number? Be as precise in your answer as you can. (h)
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