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

Short Answer

Expert verified
The binomial coefficients count the same set of selections from \( n \), representing subset and complement. Thus, \( \binom{n}{k} = \binom{n}{n-k} \).

Step by step solution

01

Understanding Binomial Coefficients

The expression \( \binom{n}{k} \) is known as a binomial coefficient and it represents the number of ways to choose \( k \) elements from a set of \( n \) elements.
02

Reformulating the Problem

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

Applying the Definition of Binomial Coefficients

The formula for a binomial coefficient is given by: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Similarly, \[ \binom{n}{n-k} = \frac{n!}{(n-k)!k!} \].
04

Comparing Both Expressions

Notice that both expressions \( \frac{n!}{k!(n-k)!} \) and \( \frac{n!}{(n-k)!k!} \) are mathematically identical, as multiplication is commutative and the expressions are equivalent.
05

Interpretation in Terms of Subsets

Choosing \( k \) elements to create a subset from a set of \( n \) elements is equivalent to selecting \( n-k \) elements to form the complement subset, leaving out those chosen. Thus, both expressions count the same selection.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of elements within a set. It is a fundamental tool in probability, algebra, and geometry. When we dive into combinatorics, we encounter concepts like permutations, combinations, and the Pascal's Triangle.

One of the primary objectives of combinatorics is to count the number of ways objects can be arranged or selected. This counting process often involves using formulas or principles devised to simplify these calculations. With combinatorics, we can solve problems related to classifying and selecting items efficiently.
  • Understanding different arrangements: How many ways can objects be ordered?
  • Counting selections: How can we choose a subset from a group?
  • Using the right tools: Employing binomial coefficients, permutations, and other formulas
Combinatorics is not only about counting; it also provides a way to understand phenomena in fields such as computer science, logic, and biological sciences. Once you grasp the basic principles, you can apply them to solve more complex problems.
Subset Selection
Subset selection is the principle of choosing a subset from a larger set. Suppose we have a set of size \( n \), and we want to know how many different subsets of size \( k \) can be formed from it. This question is precisely what binomial coefficients answer.

For example, if you have a set of five fruits and want to select three, subset selection helps determine how many different groups of three fruits you can make. Using combinatorial tools, particularly binomial coefficients, we calculate these numbers easily.
  • Key formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
  • Choosing subsets: \( k \) items out of \( n \).
In the context of the exercise, subset selection shows that choosing \( k \) items from \( n \) items is identical to choosing the remaining \( n - k \) items. This symmetry reflects in the formula: \( \binom{n}{k} = \binom{n}{n-k} \). Every time you make a selection, there's a mirror operation that selects the rest.
Binomial Theorem
The binomial theorem is a fundamental concept closely linked with binomial coefficients. It provides a powerful algebraic tool to expand expressions that are raised to a power, such as \( (x + y)^n \). The coefficients appearing in the expansion are the same binomial coefficients we've been discussing.

The theorem states:\[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k \]This formula expresses the expansion of a binomial raised to a power in terms of a sum of products involving binomial coefficients.
  • Expansion of powers: Breaking down \( (x + y)^n \) into manageable terms
  • Symmetry in coefficients: Reflects the equivalence between different subset selections
In practice, the binomial theorem is not only valuable in mathematics but also in probability and statistics, helping to solve problems involving distributions and likelihoods. The symmetry property discussed in the exercise, \( \binom{n}{k} = \binom{n}{n-k} \), is part of this broader framework and illustrates the theorem's inherent beauty and utility.

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Most popular questions from this chapter

Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, \(60 \%\) have an emergency locator, whereas \(90 \%\) of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. a. If it has an emergency locator, what is the probability that it will not be discovered? b. If it does not have an emergency locator, what is the probability that it will be discovered?

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost \(\\}\). Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost, given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Mgmt., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that \(16 \%\) of all ticks in a certain location carry Lyme disease, \(10 \%\) carry HGE, and \(10 \%\) of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

The proportions of blood phenotypes in the U.S. population are as follows: \(\begin{array}{lccc}\mathrm{A} & \mathrm{B} & \mathrm{AB} & \mathrm{O} \\ .40 & .11 & .04 & .45\end{array}\) Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? What is the probability that the phenotypes of two randomly selected individuals match?

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive vehicles pass or fail independently of one another, calculate the following probabilities: a. \(P\) (all of the next three vehicles inspected pass) b. \(P\) (at least one of the next three inspected fails) c. \(P\) (exactly one of the next three inspected passes) d. \(P\) (at most one of the next three vehicles inspected passes) e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

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