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Chapter 7: Problem 19

Evaluate the given expression. $$ C(n, 2) $$

Short Answer

Expert verified
To evaluate the given expression \(C(n, 2)\), we use the combination formula: \(C(n, k) = \frac{n!}{k!(n-k)!}\). In this case, we have n = n and k = 2. Plugging in these values and simplifying, we get the final expression: \(C(n, 2) = \frac{n(n-1)}{2}\).

Step by step solution

01

Identify the values of n and k

In the given expression, we want to find C(n, 2), which means we are given n items in a set and choosing 2 items. So, n = n and k = 2.
02

Plug the values of n and k into the combination formula

Now, we'll plug in the values of n and k into the combination formula: $$ C(n, 2) = \frac{n!}{2!(n-2)!} $$
03

Simplify the expression

To simplify the expression, we need to simplify the factorials and the division. First, let's simplify the factorials: $$ C(n, 2) = \frac{n(n-1)(n-2)!}{2!(n-2)!} $$ Now, we can cancel out the common factors (n-2)!: $$ C(n, 2) = \frac{n(n-1)}{2!} $$ Finally, we'll simplify the 2!: $$ C(n, 2) = \frac{n(n-1)}{2} $$ So the expression simplifies to: $$ C(n, 2) = \frac{n(n-1)}{2} $$

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Key Concepts

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

Combination Formula
Have you ever wondered how we calculate the number of ways to choose items from a group? That's where the combination formula comes in! This formula helps us find how many ways we can select "k" items from a total of "n" items without considering the order of selection.
The basic idea is that the order in which we choose items does not matter. Think of it like choosing two toppings for your pizza from a list of five; it doesn't matter if you pick cheese before pepperoni or vice versa, the combination of toppings is the same.
  • The general formula is expressed as \( C(n, k) = \frac{n!}{k!(n-k)!} \).
  • "n" is the total number of items, and "k" is the number of items to choose.
  • The formula accounts for duplicate arrangements by dividing by \( k! \), and the unused items by \((n-k)!\).
This approach ensures efficient calculation of combinations, helping us deal with probability and statistics problems with ease.
Factorials
Factorials are a fundamental concept when working with combinations. They are used to calculate the total number of ways to arrange a set of items. It is represented by the symbol "!".
For example, \( n! \) means multiplying all whole numbers from 1 to "n" together. If you picture a line of people and want to know the number of ways you can arrange them, that's where factorials shine.
  • The factorial of 5, written as \( 5! \), equals \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
  • Special case: \( 0! = 1 \). This is a key identity, ensuring the math works out even when selecting nothing.
Factorials grow very quickly as numbers increase and form the backbone of many mathematical expressions, including combinations.
Simplifying Expressions
The process of simplifying expressions, especially in combinations, is crucial for making calculations more manageable. In our example, simplifying \( C(n, 2) \) involves breaking down the expression into more straightforward parts.
First, we rearrange the expression \( \frac{n!}{2!(n-2)!} \) by expanding it step-by-step. This allows us to cancel common terms, thereby simplifying the factorial expressions, like replacing \( \frac{n(n-1)(n-2)!}{2!(n-2)!} \) with \( \frac{n(n-1)}{2!} \).
  • Cancel common terms: If both the numerator and denominator have the same factors, they can be eliminated to simplify.
  • Special mentions: Iin \( C(n, 2) = \frac{n(n-1)}{2} \), \( 2! = 2 \), reducing complexity and avoiding large numbers.
Understanding how to streamline expressions not only saves time but also minimizes calculation errors. This skill is handy in solving real-world problems where combinations are involved.

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Most popular questions from this chapter

A poll was conducted among 250 residents of a certain city regarding tougher gun-control laws. The results of the poll are shown in the table: $$ \begin{array}{lccccc} \hline & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Handgun } \end{array} & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Rifle } \end{array} & \begin{array}{c} \text { Own a } \\ \text { Handgun } \\ \text { and a Rifle } \end{array} & \begin{array}{c} \text { Own } \\ \text { Neither } \end{array} & \text { Total } \\ \hline \text { Favor } & & & & & \\ \text { Tougher Laws } & 0 & 12 & 0 & 138 & 150 \\ \hline \begin{array}{l} \text { Oppose } \\ \text { Tougher Laws } \end{array} & 58 & 5 & 25 & 0 & 88 \\ \hline \text { No } & & & & & \\ \text { Opinion } & 0 & 0 & 0 & 12 & 12 \\ \hline \text { Total } & 58 & 17 & 25 & 150 & 250 \\ \hline \end{array} $$ If one of the participants in this poll is selected at random, what is the probability that he or she a. Favors tougher gun-control laws? b. Owns a handgun? c. Owns a handgun but not a rifle? d. Favors tougher gun-control laws and does not own a handgun?

In a television game show, the winner is asked to select three prizes from five different prizes, \(A, B\), \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E} .\) a. Describe a sample space of possible outcomes (order is not important). b. How many points are there in the sample space corresponding to a selection that includes A? c. How many points are there in the sample space corresponding to a selection that includes \(\mathrm{A}\) and \(\mathrm{B}\) ? d. How many points are there in the sample space corresponding to a selection that includes either \(\mathrm{A}\) or \(\mathrm{B}\) ?

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$ \begin{array}{lccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

List the simple events associated with each experiment. Data concerning durable goods orders are obtained each month by an economist. A record is kept for a 1 -yr period of any increase \((i)\), decrease \((d)\), or unchanged movement \((u)\) in the number of durable goods orders for each month as compared with the number of such orders in the same month of the previous year.

The grade distribution for a certain class is shown in the following table. Find the probability distribution associated with these data. $$ \begin{array}{lccccc} \hline \text { Grade } & \text { A } & \text { B } & \text { C } & \text { D } & \text { F } \\ \hline \text { Frequency of } & & & & & \\ \text { Occurrence } & 4 & 10 & 18 & 6 & 2 \\ \hline \end{array} $$
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