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From the symmetry of the binomial coefficients, it is not too hard to see that when \(n\) is an odd number, the number of subsets of \(\\{1,2, \ldots, n\\}\) of odd size equals the number of subsets of \(\\{1,2, \ldots, n\\}\) of even size. Is it true that when \(n\) is even the number of subsets of \(\\{1,2, \ldots, n\\}\) of even size equals the number of subsets of odd size? Why or why not? (h)

Short Answer

Expert verified
Yes, the number of subsets of even size equals those of odd size due to the symmetry of binomial coefficients.

Step by step solution

01

Understanding Binomial Coefficients

Binomial coefficients, denoted as \(\binom{n}{k}\), represent the number of ways to choose \(k\) elements from a set of \(n\) elements. For a set \(\{1, 2, \ldots, n\}\), the total number of subsets is \(2^n\), divided into those with an even number of elements and those with an odd number of elements.
02

Total Subsets Calculation

The total number of subsets of a set with \(n\) elements is \(2^n\). This includes subsets of all sizes from 0 to \(n\).
03

Equal Division of Subsets

For any integer \(n\), the number of subsets of even size and the number of subsets of odd size will sum to \(2^n\). Analyzing these subsets can be done by looking at the binomial coefficients. When \(n\) is odd, it's evident that each binomial coefficient \(\binom{n}{k}\) for odd \(k\) matches an even \(k\) one.
04

Symmetry of Binomial Coefficients

The symmetry of binomial coefficients indicates that \(\binom{n}{k} = \binom{n}{n-k}\). This property means that choosing \(k\) elements from \(\binom{n}{k}\) is the same as choosing \((n-k)\) elements, reflecting subsets of even and odd sizes around \(n/2\). When \(n\) is even, \(\binom{n}{k}\) still produces symmetrical pairs.
05

Parity Analysis of Binomial Coefficients

To analyze for even \(n\), note that \(\frac{n}{2}\) is an integer. Since \(\binom{n}{k}\) pairs every binomial class of odd size with a binomial class of even size due to the symmetry \(\binom{n}{k} = \binom{n}{n-k}\), it ensures an equal count when they sum up.
06

Conclusion

Thus, irrespective of being even or odd, due to the symmetrical nature of the binomial coefficients and the equal division by their parity, the subsets of even and odd sizes remain equal for any integer \(n\).

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Key Concepts

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

Symmetry of Binomial Coefficients
Binomial coefficients are represented as \( \binom{n}{k} \). This notation stands for the number of ways to choose \( k \) elements from a set of \( n \) elements. One important property of binomial coefficients is their symmetry. This symmetry can be written as \( \binom{n}{k} = \binom{n}{n-k} \), which means that choosing \( k \) elements is the same as leaving out \( k \) elements and choosing the remaining \( n - k \) elements.Consider a set \( \{1, 2, \ldots, n\} \). Because of the symmetry, each subset size \( k \) has a counterpart \( n - k \). If \( k \) is odd, then \( n - k \) is even, and vice versa. This pairing is crucial for analyzing the distribution of subset sizes when counting subsets of even and odd sizes.
Parity Analysis
Parity analysis involves examining the evenness and oddness of binomial coefficients. The symmetry property \( \binom{n}{k} = \binom{n}{n-k} \) means we can pair up subsets with odd and even sizes. When \( n \) is odd, each binomial coefficient for odd \( k \) matches an even \( k \). This ensures equal numbers of subsets with odd and even sizes. When \( n \) is even, the same pairing occurs because \( \frac{n}{2} \) is an integer and each \( k \) from \( 1 \) to \( \frac{n}{2} \) has a corresponding \( n-k \). Thus, for any \( n \), the symmetry ensures that there are equal numbers of subsets of odd and even sizes.
Subset Counting
To count subsets of a set \( \{1, 2, \ldots, n\} \), consider the total number of subsets, which is \( 2^n \). This includes subsets of all sizes from 0 to \( n \).Since subsets are either even-sized or odd-sized, we can split \( 2^n \) into two parts: the number of even-sized subsets and the number of odd-sized subsets. The equation becomes \[ \text{Number of even-sized subsets} + \text{Number of odd-sized subsets} = 2^n \] Analysis using binomial coefficients shows that the number of subsets with a particular size is \( \binom{n}{k} \). Due to the symmetry property and parity pairing, regardless of \( n \) being odd or even, the number of subsets of even size equals the number of subsets of odd size.Thus, it's true for both odd and even \( n \) that the number of subsets of even size equals the number of subsets of odd size.

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Most popular questions from this chapter

Now some number \(n\) of schools are going to send their baseball teams to a tournament, and each team must play each other team exactly once. Let us think of the teams as numbered 1 through \(n\). (a) How many games does team 1 have to play in? (b) How many games, other than the one with team \(1,\) does team two have to play in? (c) How many games, other than those with the first \(i-1\) teams, does team \(i\) have to play in? (d) In terms of your answers to the previous parts of this problem, what is the total number of games that must be played?

Two sets are said to be disjoint if they have no elements in common. For example, \(\\{1,3,12\\}\) and \(\\{6,4,8,2\\}\) are disjoint, but \(\\{1,3,12\\}\) and \(\\{3,5,7\\}\) are not. Three or more sets are said to be mutually disjoint if no two of them have any elements in common. What can you say about the size of the union of a finite number of finite (mutually) disjoint sets? Does this have anything to do with any of the previous problems?

Your formula for the Catalan Number can be expressed as a binomial coefficient divided by an integer. Whenever we have a formula that calls for division by an integer, an ideal combinatorial explanation of the formula is one that uses the quotient principle. The purpose of this problem is to find such an explanation using diagonal lattice paths. \({ }^{a}\) A diagonal lattice path that never goes below the \(y\) -coordinate of its first point is called a Dyck Path. We will call a Dyck Path from (0,0) to \((2 n, 0)\) a Catalan Path of length \(2 n\). Thus the number of Catalan Paths of length \(2 n\) is the Catalan Number \(C_{n}\) (a) If a Dyck Path has \(n\) steps (each an upstep or downstep), why do the first \(k\) steps form a Dyck Path for each nonnegative \(k \leq n ?\) (b) Thought of as a curve in the plane, a diagonal lattice path can have many local maxima and minima, and can have several absolute maxima and minima, that is, several highest points and several lowest points. What is the \(y\) -coordinate of an absolute minimum point of a Dyck Path starting at (0,0)\(?\) Explain why a Dyck Path whose rightmost absolute minimum point is its last point is a Catalan Path. (h) (c) Let \(D\) be the set of all diagonal lattice paths from (0,0) to \((2 n, 0)\). (Thus these paths can go below the \(x\) -axis.) Suppose we partition \(D\) by letting \(B_{i}\) be the set of lattice paths in \(D\) that have \(i\) upsteps (perhaps mixed with some downsteps) following the last absolute minimum. How many blocks does this partition have? Give a succinct description of the block \(B_{0} \cdot(\mathrm{h})\) (d) How many upsteps are in a Catalan Path? (e) We are going to give a bijection between the set of Catalan Paths and the block \(B_{i}\) for each \(i\) between 1 and \(n\). For now, suppose the value of \(i,\) while unknown, is fixed. We take a Catalan path and break it into three pieces. The piece \(F\) (for "front") consists of all steps before the \(i\) th upstep in the Catalan path. The piece \(U\) (for "up") consists of the \(i\) th upstep. The piece \(B\) (for "back") is the portion of the path that follows the \(i\) th upstep. Thus we can think of the path as \(F U B\). Show that the function that takes \(F U B\) to \(B U F\) is a bijection from the set of Catalan Paths onto the block \(B_{i}\) of the partition. (Notice that \(B\) UF can go below the \(x\) axis.) \((\mathrm{h})\) (f) Explain how you have just given another proof of the formula for the Catalan Numbers.

A basketball team has 12 players. However, only five players play at any given time during a game. (a) In how may ways may the coach choose the five players? (b) To be more realistic, the five players playing a game normally consist of two guards, two forwards, and one center. If there are five guards, four forwards, and three centers on the team, in how many ways can the coach choose two guards, two forwards, and one center? (h) (c) What if one of the centers is equally skilled at playing forward? (h)

How many subsets does a set \(S\) with \(n\) elements have? (b)

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