Help with a question about collinear points

  • Thread starter Thread starter mathstudent88
  • Start date Start date
  • Tags Tags
    Points
Click For Summary
SUMMARY

The discussion centers on calculating the number of lines that can be formed by two points selected from five distinct points, where no three points are collinear. The correct approach involves using combinations, specifically C(5,2), which calculates to 10 lines, not 60 as initially miscalculated. The formula for combinations is confirmed as C(n, k) = n! / (k!(n-k)!), leading to the conclusion that the correct number of lines is 10.

PREREQUISITES
  • Understanding of combinations in combinatorial mathematics
  • Familiarity with factorial notation and calculations
  • Basic knowledge of geometric principles regarding points and lines
  • Ability to interpret mathematical notation and expressions
NEXT STEPS
  • Study the concept of combinations in depth, focusing on C(n, k) calculations
  • Explore advanced combinatorial problems involving points and lines
  • Learn about geometric properties of collinear and non-collinear points
  • Practice solving problems related to permutations and combinations
USEFUL FOR

Students in mathematics, educators teaching combinatorial concepts, and anyone interested in geometric properties and calculations involving points and lines.

mathstudent88
Messages
25
Reaction score
0
Let P1, P2, ..., P5 be five points, no three of which are collinear. How many lines contain two of these five points?


How I thought about solving this was by way of combinations.

C5,2= (5*4*3*2)/2 = 60.


So there would be 60 lines that contain 2 of the 5 points. Is this a good way to approach this problem or is there a better way?

Thank you for your help!
 
Last edited:
Physics news on Phys.org
Welcome to PF!

Hi mathstudent88! Welcome to PF! :smile:

(hmm … 60 is a very large number … :rolleyes:)
mathstudent88 said:
Let P1, P2, ..., P5 be five points, no three of which are collinear. How many lines contain two of these five points?

How I thought about solving this was by way of combinations.

C5,2= (5*4*3*2)/2 = 60.

So there would be 60 lines that contain 2 of the 5 points. Is this a good way to approach this problem or is there a better way?

No … combinations is exactly the right way! :smile:

but … C5,2 = 5!/2!(5-2)! = … ? :wink:
 
haha I forgot all about the (5-2)! part... thank you for pointing it out for me!

the answer is 10.


Thanks again!
 

Similar threads

Prove that five points are collinear
  • · Replies 5 ·
Replies
5
Views
3K
Collinear Points -- Ways to determine if points are collinear
  • · Replies 20 ·
Replies
20
Views
3K
Does there exist a 2x2 non-singular matrix with only one 1d eigenspace?
  • · Replies 2 ·
Replies
2
Views
2K
Finding the point on the line of intersection between planes
  • · Replies 5 ·
Replies
5
Views
2K
Understanding Fixed Points in Hamiltonian Systems
  • · Replies 2 ·
Replies
2
Views
2K
Constructing cubic spline interpolation polynomials
  • · Replies 4 ·
Replies
4
Views
2K
Number of lines equidistant from four points on a plane
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Finding Collinear Points Given 1 point and 2 lines
  • · Replies 17 ·
Replies
17
Views
5K
Projective geometry question: 4 points no 3 on a line
  • · Replies 3 ·
Replies
3
Views
3K
Question on definite integral and inflection points
  • · Replies 2 ·
Replies
2
Views
2K