Question 3

Geometry Level 2

Seven points are marked on the circumference of a circle. How many different chords can be drawn by connecting two of these seven points?

The answer is 21.

5 solutions

Trevor B.
May 25, 2014

A chord can start at any of the $7$ points and end at any of the other $6$ points. However, we have to divide by two to avoid double counting, so the answer is $\dfrac{7\times6}{2}=\boxed{21}$

Alternatively, the number of chords that can be constructed is $\dbinom{7}{2}=\dfrac{7!}{5!\times2!}=\dfrac{7\times6}{2}=\boxed{21}$

the ans is 7c2=7!/{(7-2)! x 2!}................. here 7 is the number of points & 2 is on which way they are joined.....the formula is: n!/{(n-k)! x k!}

Rayhan Ferdous Faisal - 7 years ago

$nCk=\dfrac{n!}{k!\times(n-k)!}$ is denoted by $\dbinom{n}{k}$

Trevor B. - 7 years ago

there are 7points from the 1st point 6chords are possible so in 2nd 5chords are possible similarly 6+5+4+3+2+1=21

Sauvik Connors - 7 years ago

6+5+4+3+2+1 = 1+2+.....+n where n is 6 so n(n+1)/2 = 6*7/2 so 21

Ansh Bhatt - 7 years ago

why to do by such a long method. Do just 6+5+4+3+2+1+0

Anuj Shikarkhane - 6 years, 11 months ago

Hanuman Bajrangbali
Nov 17, 2016

This is a case of NcR or N 'choose' R, where it is defined as n! /r!×(n-r)! In this case it would be 7!/2!×(7-2)! Which equals 21

Ramiel To-ong
Jun 12, 2015

the formula for that is n(n-1)/2 where n = number of points in the circumference

Krishna Sharma
Oct 9, 2014

Hey @trevor can you help me with this ?

Aryan Jakhar
Jul 5, 2014

$To\quad Find\quad The\quad Number\quad Of\quad Lines\quad Between\quad any\quad Number\quad Of\quad Non-CoLinear\quad Points.\\ For\quad n\quad Point\quad The\quad No.\quad Of\quad Lines\quad Will\quad Be\quad \cfrac { n(n-1) }{ 2 } .\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 7\times 6 }{ 2 } =21$

Question 3

The answer is 21.

5 solutions

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