Problem 24 - Algebra for College Students

Question

Solve each problem. See Examples \(3-5\) How many distinct triangles are determined by 5 points lying on a circle, where the vertices of each triangle are chosen from the 5 points?

Step-by-Step Solution

Answer

10 distinct triangles.

Step 1: Understanding the Problem

We have 5 points on a circle. We need to determine how many distinct triangles can be formed using 3 of these points as vertices.

Step 2: Recognize it as a Combination Problem

To find the number of ways to choose 3 points out of 5, we use the concept of combinations, denoted by \( nCk \). In this case, \( n = 5 \) and \( k = 3 \).

Step 3: Use the Combination Formula

The combination formula is given by: \[ nCk = \frac{n!}{k!(n-k)!} \]

Step 4: Substitute Values into the Formula

Substitute \( n = 5 \) and \( k = 3 \) into the combination formula: \[ 5C3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} \]

Step 5: Simplify the Expression

Simplify the factorials in the equation: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120, \quad 3! = 3 \times 2 \times 1 = 6, \quad 2! = 2 \times 1 = 2 \] \[ 5C3 = \frac{120}{6 \times 2} = \frac{120}{12} = 10 \]

Step 6: Result Interpretation

There are 10 distinct ways to choose 3 points out of 5 to form a triangle on the circle.

Key Concepts

combinations in mathematics

In mathematics, combinations are a way to select items from a larger pool where the order of selection does not matter. For example, when forming a committee from a group of people, you are often interested in the different possible groups, not in the order they are chosen.
In our exercise, we are dealing with how to choose 3 points from 5 points on a circle to form a triangle. This is a perfect fit for combinations because the order in which the points are chosen does not affect the triangle formed.
The combination formula is: \(\binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n \) is the total number of items, and \( k \) is the number of items to choose. Using this formula, we can calculate the number of ways to select any 3 points from 5.

triangle formation

Forming triangles from a set of points requires understanding how many unique sets of points can create the triangle.
A triangle needs exactly three vertices. In this problem, each triangle is defined by choosing 3 out of 5 points. When points are on a circle, every group of three unique points will always form a triangle. Because of this circular arrangement, there are no collinear points that could invalidate a triangle.
So in the context of this exercise, once you know how many combinations of three points you can select from five (which is 10), you also know the number of unique triangles.

factorials in permutations

Factorials are a fundamental concept when dealing with permutations and combinations. The factorial of a number \( n! \) is the product of all positive integers less than or equal to \( n \).
In combination problems, factorials help us count the number of ways to arrange a set of items. For example, in this exercise's simplified combination formula \(\binom{5}{3} = \frac{5!}{3!2!}\), we calculate 5! (5 factorial), 3! and 2!. This breaks down to: \( 5! = 120, 3! = 6, \) and \(2! = 2\).
By plugging these into the formula, you can see how factorials help simplify and solve the problem. This use of factorials ensures the correct accounting for all possible arrangements of the items without considering the order.