Problem 43 - Algebra for College Students

Question

How many different line segments are determined by five points? By six points? By seven points? By \(n\) points? \(10 ; 15 ; 21 ; \frac{n(n-1)}{2}\)

Step-by-Step Solution

Answer

For 5: 10; For 6: 15; For 7: 21; Formula: \(\frac{n(n-1)}{2}\).

Step 1: Understanding the Problem

We are tasked to find how many unique line segments can be formed from a set number of points. In this context, a line segment is defined by two endpoints. For each unique pair of points, we have a distinct line segment.

Step 2: Formula Derivation

The number of line segments from a given number of points is equivalent to choosing 2 points out of these points to form a line segment. This is a common combinatorics problem that can be solved using the binomial coefficient formula: \( \binom{n}{2} \).

Step 3: Substituting for Five Points

Using the formula derived, substitute \( n = 5 \) in the formula \( \binom{n}{2} \):\[ \binom{5}{2} = \frac{5 \times 4}{2} = 10 \]Thus, there are 10 unique line segments with five points.

Step 4: Substituting for Six Points

Substitute \( n = 6 \) in the formula \( \binom{n}{2} \):\[ \binom{6}{2} = \frac{6 \times 5}{2} = 15 \]Therefore, there are 15 unique line segments with six points.

Step 5: Substituting for Seven Points

Substitute \( n = 7 \) in the formula \( \binom{n}{2} \):\[ \binom{7}{2} = \frac{7 \times 6}{2} = 21 \]Hence, there are 21 unique line segments with seven points.

Step 6: Generalizing the Solution

The result for \( n \) points follows a pattern derived by the formula:\[ \binom{n}{2} = \frac{n(n-1)}{2} \]This formula computes the number of unique line segments that can be drawn between any set of \( n \) points.

Key Concepts

Line Segments

Line segments are fundamental elements in geometry, defined by two endpoints on a plane. Unlike a line, which extends infinitely in both directions, a line segment has a fixed length, starting at one endpoint and ending at another.
Understanding line segments is essential because they serve as the building blocks for various geometrical constructions. In this exercise, line segments are determined by selecting any two points from a given set.
Each pair of points forms one unique line segment, meaning the order in which the points are chosen does not affect the segment's identity. This property plays a crucial role in determining the total count of line segments that can be constructed from a set of points.

Binomial Coefficient

The binomial coefficient is a key concept in combinatorics, used to calculate the number of ways to choose a set number of items from a larger collection, where the order does not matter.
This coefficient is expressed as \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items being chosen.

• For example, \( \binom{n}{2} \) expresses the number of ways to choose two items from \( n \) items.
• The formula for the binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

This math notation is pivotal when calculating combinations, such as how many line segments can be formed from multiple points. In this exercise, we specifically use \( \binom{n}{2} \) since we're creating pairs of points.

Point Selection

Point selection is the process of choosing specific points from a larger set. In the context of this exercise, selecting points is about forming lines and is governed by combinations rather than permutations.
When selecting points:

• The order in which points are selected does not matter, meaning \( A, B \) is the same as \( B, A \).
• Each unique pair of points results in a distinct line segment.

The primary goal here is to determine all possible combinations of point pairs because each combination forms a line segment. Mathematically, this is handled by the binomial coefficient \( \binom{n}{2} \) formula, which calculates the total number of unique pairs from \( n \) points.

Combinatorial Formulas

Combinatorial formulas, such as the ones used in this exercise, help solve problems regarding the arrangement or selection of objects. These formulas play a significant role in simplifying complex counting problems.
The formula \( \binom{n}{2} = \frac{n(n-1)}{2} \) is instrumental in our case. It calculates the number of unique pairs that can be selected from a set of \( n \) items.

• This formula derives from the understanding that each point can connect to every other point exactly once.
• By reducing the factorial terms in the generalized binomial formula, we arrive at this convenient expression for pairs.

Using combinatorial formulas, we efficiently solve problems that otherwise demand cumbersome and exhaustive counting. They provide a structured way to approach arrangement and selection tasks, like determining the number of line segments from given points in this exercise.