Problem 7
Question
Find the number of lines that can be drawn using 10 distinct points, no three being collinear.
Step-by-Step Solution
Verified Answer
The number of lines that can be drawn using 10 distinct points, no three being collinear is 45, which can be calculated using the combination formula: \[C(10, 2) = \frac{10!}{2!(10 - 2)!} = 45\].
1Step 1: Define the problem in terms of combinations
In this problem, we need to find the number of lines that can be formed using 10 distinct points, no three being collinear. We can frame this problem as choosing 2 points out of the 10 to form a line, as every pair of points uniquely determines a line. So essentially, we need to find the number of combinations of choosing two points out of the 10 points.
2Step 2: Use the combination formula
The combination formula is given by \[C(n, k) = \frac{n!}{k!(n - k)!}\], where \(n\) represents the total number of items, and \(k\) represents the number of items you want to choose from that total.
In our problem, we have a total of 10 points, so \(n = 10\), and we want to choose 2 points out of these 10 to form a line, so \(k = 2\).
3Step 3: Calculate the number of lines
Using the combination formula, we can calculate the number of lines that can be formed using 10 distinct points as follows:
\[C(10, 2) = \frac{10!}{2!(10 - 2)!} = \frac{10!}{2!8!}\]
Now, calculate the factorials:
\[= \frac{10 \times 9 \times 8!}{2! \times 8!}\]
We can see that the 8! in the numerator and denominator will cancel each other. Now, calculate the remaining terms:
\[= \frac{10 \times 9}{2}\]
\[= 5 \times 9\]
\[= 45\]
So there are 45 lines that can be drawn using the 10 distinct points, no three being collinear.
Key Concepts
Combinatorial MathematicsDiscrete MathematicsNon-Collinear Points
Combinatorial Mathematics
In combinatorial mathematics, we deal with the art of counting and arranging objects following certain rules without actually enumerating them individually. The concept is central to discrete mathematics, as it involves understanding how to count different configurations effectively. For example, when trying to determine the number of lines that can be drawn from a given set of points, we use combinations because each line is defined by a distinct pair of points.
The formula for combinations is expressed as \[C(n, k) = \frac{n!}{k!(n - k)!}\], where \(n\) is the total number of items, and \(k\) is the number of items to choose. Combinatorial mathematics comes into play when we want to select a subset from a larger set, counting the number of ways to do this without care for the order of selection, which is exactly what we have done in our exercise solution. The simplicity and elegance of combinatorial formulas allow us to solve complex problems with relative ease.
The formula for combinations is expressed as \[C(n, k) = \frac{n!}{k!(n - k)!}\], where \(n\) is the total number of items, and \(k\) is the number of items to choose. Combinatorial mathematics comes into play when we want to select a subset from a larger set, counting the number of ways to do this without care for the order of selection, which is exactly what we have done in our exercise solution. The simplicity and elegance of combinatorial formulas allow us to solve complex problems with relative ease.
Discrete Mathematics
Discrete mathematics is a branch of mathematics that deals with distinct and separate values, usually integers. It includes a variety of subfields such as graph theory, combinatorics, algorithms, and more. In the context of our exercise, discrete mathematics is pertinent as we tackle problems using a finite set of points. Each point is considered an individual entity.
Discrete mathematics encourages a step-by-step approach to problem-solving. Sequentially, we defined our problem, applied the correct formula for combinations, and calculated the result. By focusing on discrete elements (the points) rather than a continuous spectrum, discrete mathematics makes it feasible to count and calculate elements like individual lines in our problem scenario. It's all about breaking the problem down into manageable, quantifiable pieces, which makes for more accessible and understandable solutions.
Discrete mathematics encourages a step-by-step approach to problem-solving. Sequentially, we defined our problem, applied the correct formula for combinations, and calculated the result. By focusing on discrete elements (the points) rather than a continuous spectrum, discrete mathematics makes it feasible to count and calculate elements like individual lines in our problem scenario. It's all about breaking the problem down into manageable, quantifiable pieces, which makes for more accessible and understandable solutions.
Non-Collinear Points
Non-collinear points are a set of points that do not all lie on the same straight line. This is an essential concept when considering the number of lines that can be formed from distinct points without any three points being collinear. In our exercise, it is specified that the 10 points are non-collinear; this implies that no line can intersect more than two points at a time.
Understanding non-collinearity is crucial since it directly impacts the count of possible lines. If three points were collinear, they would define only one line rather than the several combinations possible when points are non-collinear. In discrete mathematics, when modeling certain situations like network connections or pathways, the notion of non-collinear points often naturally arises, and it ensures a higher complexity and a more extensive set of connections, much like the problem we have worked on with the 10 distinct points.
Understanding non-collinearity is crucial since it directly impacts the count of possible lines. If three points were collinear, they would define only one line rather than the several combinations possible when points are non-collinear. In discrete mathematics, when modeling certain situations like network connections or pathways, the notion of non-collinear points often naturally arises, and it ensures a higher complexity and a more extensive set of connections, much like the problem we have worked on with the 10 distinct points.
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