Problem 13 - A Graphical Approach to College Algebra

Question

Evaluate the following. In Exercises 17 and \(18,\) express the answer in terms of \(n .\) Do not use a calculator. $$\left(\begin{array}{c}100 \\\2\end{array}\right)$$

Step-by-Step Solution

Answer

The combination \( \binom{100}{2} \) evaluates to 4950.

Step 1: Understanding Combinations

In this exercise, we need to calculate a combination, denoted as \( \binom{100}{2} \). This represents the number of ways to choose 2 items from a set of 100, without considering the order. It is given by the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

Step 2: Substituting Values into Formula

We substitute \( n = 100 \) and \( k = 2 \) into the combination formula: \[ \binom{100}{2} = \frac{100!}{2!(100-2)!} = \frac{100!}{2! \, 98!} \].

Step 3: Simplifying the Factorials

To simplify, note that \( 100! = 100 \times 99 \times 98! \). So the expression becomes: \[ \frac{100 \times 99 \times 98!}{2! \times 98!} \]. The 98! terms cancel out.

Step 4: Calculating Remaining Terms

Now we have: \[ \frac{100 \times 99}{2!} = \frac{100 \times 99}{2 \times 1} = \frac{100 \times 99}{2} = 50 \times 99 \].

Step 5: Final Multiplication

Perform the final multiplication: \( 50 \times 99 = 4950 \). This is the number of ways to choose 2 items from 100.

Key Concepts

Factorials

Factorials are a fundamental concept in mathematics, especially when dealing with permutations and combinations. The factorial of a number \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). This concept is very useful for counting and arranging objects.

When dealing with large numbers, factorials can be quite extensive, which is why factorial notation is a compact way of representing these calculations. In our problem, to find \( \binom{100}{2} \), we need to compute \( 100! \). However, since we are dividing by \( (100-2)! \), most of the factorial terms cancel out, simplifying our work. Only the necessary terms, in this case, \( 100 \times 99 \), are calculated directly.

Binomial Coefficient

The binomial coefficient is a key concept in combinatorics, represented as \( \binom{n}{k} \), and is used to determine the number of ways to choose \( k \) items from a total of \( n \) without considering the order. This is also known as a combination. The formula for calculating the binomial coefficient is:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

In this formula, \( n! \) is the factorial of \( n \), \( k! \) is the factorial of \( k \), and \( (n-k)! \) is the factorial of the difference between \( n \) and \( k \).

Using our given example, \( \binom{100}{2} = \frac{100!}{2!(100-2)!} \). After simplifying the factorials, this results in calculating \( \frac{100 \times 99}{2 \times 1} \). The binomial coefficient formula elegantly reduces complex decision-making problems into manageable calculations.

Discrete Mathematics

Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. It includes topics such as graphs, integers, and statements in logic. Combinations, like those calculated using the binomial coefficient, play a huge role in this field because they allow us to solve a wide range of problems involving discrete objects.

In combinatorics, which is a branch of discrete mathematics, we often tackle the challenge of counting distinct arrangements or selections from a larger set. This is integral to computer science, cryptography, and optimization problems.

Our exercise involves a combination problem, a typical issue in discrete mathematics, requiring the computation of how many ways there are to select a subset of items from a given set, highlighting the importance of understanding factorials and the binomial coefficient.