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Show that \(\left(\begin{array}{c}n \\\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)\). Give an interpretation involving subsets.

Short Answer

Expert verified
The equality holds due to symmetric properties of binomial coefficients and can be interpreted as choosing \(k\) items or equivalently not choosing \(n-k\) items.

Step by step solution

01

Understanding Binomial Coefficients

Binomial coefficients, written as \(\binom{n}{k}\), count the number of ways to choose \(k\) elements from a set of \(n\) elements. This is computed using the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
02

Formulating the Problem

We need to show that \(\binom{n}{k} = \binom{n}{n-k}\). This means the number of ways to choose \(k\) elements from \(n\) is the same as choosing \(n-k\) elements from \(n\).
03

Using Symmetric Properties of Binomial Coefficients

By the definition of binomial coefficients, we have \(\binom{n}{n-k} = \frac{n!}{(n-k)!k!}\). Notice this is identical to \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). The formulas are equivalent due to the commutative property of multiplication.
04

Conceptual Understanding with Subsets

Choosing \(k\) items from a set of \(n\) items can be viewed as choosing \(n-k\) items that you do not pick. Therefore, the number of ways to choose \(k\) items is the same as the number of ways to choose the \(n-k\) items that are left out.

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Key Concepts

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

Combinatorics
Combinatorics is the area of mathematics that deals with counting, arrangement, and combination of elements in sets. It provides tools to count the number of ways certain actions can be performed. One fundamental component in combinatorics is the concept of choosing elements from a larger set.
Consider a set of elements; combinatorics helps you determine how many ways you can select, arrange, or distribute these elements according to specific rules. This counting principle is crucial in many areas, including probability and statistics.
Binomial coefficients are a key part of combinatorics. They express the number of ways to choose a specified number of elements from a larger set, disregarding the order of selection. Recognizable by the notation \( \binom{n}{k} \), a binomial coefficient is calculated using the formula:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Here, \( n! \) ("n factorial") represents the product of all positive integers up to \( n \), and \( (n-k)! \) represents the factorial of the difference between the number of total elements and the subset size, \( k \). Understanding how these coefficients work is essential in combinatorial mathematics.
Subsets
Subsets represent smaller groups formed from a larger set by choosing some or no elements. In a given set, there can be many possible subsets. For a set with \( n \) elements, the total number of subsets, including the empty set, is \( 2^n \).
When you're working with binomial coefficients, you're effectively counting subsets. The expression \( \binom{n}{k} \) specifically counts the \( k \)-sized subsets of an \( n \)-element set.
For example, if you have a set with three elements \( \{a, b, c\} \), the possible subsets include:
  • Size 0 (empty subset): \( \{\} \)
  • Size 1: \( \{a\} \), \( \{b\} \), \( \{c\} \)
  • Size 2: \( \{a, b\} \), \( \{a, c\} \), \( \{b, c\} \)
  • Size 3: \( \{a, b, c\} \)
The binomial coefficient \( \binom{3}{2} \) equals 3 because there are three 2-element subsets: \( \{a, b\} \), \( \{a, c\} \), and \( \{b, c\} \). Understanding subsets and how they relate to binomial coefficients helps you grasp many combinatorial problems.
Symmetric Properties
Symmetric properties in the context of binomial coefficients refer to certain equalities that arise due to the mathematical structure of combinations. An especially important symmetric property is the relationship given by:
  • \( \binom{n}{k} = \binom{n}{n-k} \)
This means that choosing \( k \) elements from \( n \) is the same as choosing \( n-k \) elements to exclude.
Let's break down why this symmetry exists:
When you select \( k \) elements, there's an implicit choice of \( n-k \) elements that you are leaving out. Therefore, every combination of \( k \) elements corresponds to a complementary combination of \( n-k \) non-selected elements. These two quantities are equal because the process of selecting a subset inherently involves not selecting its complement.
This symmetric property is valuable because it simplifies combinatorial proofs and calculations, allowing us to interchange \( k \) and \( n-k \) freely in binomial coefficient expressions. Understanding symmetry helps us appreciate the inherent balance in choosing elements, either directly selecting or indirectly through exclusion.

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Most popular questions from this chapter

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for indepth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

A box in a certain supply room contains four \(40-W\) lightbulbs, five \(60-\mathrm{W}\) bulbs, and six \(75-\mathrm{W}\) bulbs. Suppose that three bulbs are randomly selected. a. What is the probability that exactly two of the selected bulbs are rated 75 W? b. What is the probability that all three of the selected bulbs have the same rating? c. What is the probability that one bulb of each type is selected? d. Suppose now that bulbs are to be selected one by one until a 75 -W bulb is found. What is the probability that it is necessary to examine at least \(\operatorname{six}\) bulbs?

Two voters, \(\mathrm{Al}\) and Bill, are each choosing between one of three candidates \(-1,2\), and \(3-\) who are running for city council. An experimental outcome specifies both Al's choice and Bill's choice, e.g. the pair \((3,2)\). a. List all elements of \(S\). b. List all outcomes in the event \(A\) that \(\mathrm{Al}\) and Bill make the same choice. c. List all outcomes in the event \(B\) that neither of them vote for candidate 2 .

Consider randomly selecting a student at a certain university, and let \(A\) denote the event that the selected individual has a Visa credit card and \(B\) be the analogous event for a MasterCard. Suppose that \(P(A)=.5, P(B)=.4\), and \(P(A \cap B)=.25\). a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event \(A \cup B\) ). b. What is the probability that the selected individual has neither type of card? c. Describe, in terms of \(A\) and \(B\), the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

Three molecules of type \(A\), three of type \(B\), three of type \(C\), and three of type \(D\) are to be linked together to form a chain molecule. One such chain molecule is \(A B C D A B C D A B C D\), and another is \(B C D D A A A B D B C C\). a. How many such chain molecules are there? [Hint: If the three A's were distinguishable from one another- \(A_{1}, A_{2}, A_{3}\)-and the \(B\) 's, \(C\) 's, and \(D\) 's were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the \(A\) 's?] b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to each other (such as in \(B B B A A A D D D C C C\) )?

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