Question

Given two points A and B, there is only one segment that you can name: AB. With three collinear points A, B, and C, there are three different segemnts that you can name: AB, AC, and BC. With five colinear points A, B, C. D, and E, how many different segments can you name?

Answer 1

Using the combinations formula C(n, k) = n! / (k!(n-k)!), with n=5 and k=2 for the five collinear points A, B, C, D, and E, there are 10 different segments that can be named.

Step-by-step explanation:

Calculating the Number of Line Segments

With five collinear points A, B, C, D, and E, we can name many different line segments. To determine the number of line segments, you can use a simple combinatorial approach: since a line segment is defined by two distinct endpoints, and you have five points to choose from, you need to calculate the number of combinations of five points taken two at a time.

The formula for combinations is C(n, k) = n! / (k!(n-k)!), where n is the total number of items to choose from, k is the number of items to choose, and ! denotes a factorial.

Using this formula, we get C(5, 2) = 5! / (2!(5-2)!) = 10. This means there are 10 different segments that you can name with five collinear points.

Answer 2

Answer: 10

Imagine you have 2 slots or boxes that are empty. They represent the possible choices for the letter you pick. For example, you can place B in slot 1 and D in slot 2. There are 5 choices for slot 1 (A,B,C,D,E) and four choices for slot 2. Why 4? Because after we pick the letter for slot 1, we have one less letter to pick from. We can't reuse that letter.

Now multiply those values 5 and 4 to get 20. There are 20 different ways to pick a pair of letters from a pool of 5 total. However, order does NOT matter because the segment AB is the same as BA. Since order doesn't matter, we are doubly counting when we shouldn't. In other words, our count is two times higher than it should be. Instead of 20 pairs, it's actually 20/2 = 10 pairs. That's why the answer is 10.

The list of 10 segments are: {AB,AC,AD,AE,BC,BD,BE,CD,CE,DE}

Side note: you can use the nCr combination formula with n = 5 and r = 2 to get the same answer.