How many triangles?

Question

Suppose you mark 10 points on a circle. How many triangles are possible that have 3 of these points as corner points? Please also include how you got them.

Answer 1

This is combination problem disguised as a geometry problem - how many ways are there to select three items from a group of ten items?

10!/3!7! = 10 * 9 * 8 / 6 = 120

Answer 2

If we marked 3 points on a circle. there would be 1 triangle.
To add 1 point on the circle which already have 3 points, we would make new triangles which 2 corners are selected from 3 old points.We may try to add a new point to infinity.

If there are (n-1) old points on the circle , How many case we can select 2 points ? It's cmbination (2,(n-1))=(n-1)(n-2)/2


add-1-point, ---- old-point, ---- new-triangles
---------------,-----------------,----------- 1 (original triangle)
-----1st------,-------- 3 ------,---------3*2/2　(4 points are there totaly)
-----2nd-----,-------- 4 ------,---------4*3/2
-----3rd-----,--------- 5 ------,---------5*4/2
・・・
----(n-3)----,-------- n-1------,-------(n-1)(n-2)/2

So we could count triangles such this.
S=1+3*2/2+4*3/2+…+(n-1)(n-2)/2
=1+(1/2){3*2+4*3+…+(n-1)(n-2)}

If n=10 then
S=1+(1/2)*(3*2+4*3+…+9*8)
=1+(1/2)*(6+12+20+30+42+56+72)
=1+(1/2)*238
=1+119
=120

Of cause,
combination(3,10)
=(10*9*8)/(3*2*1)
=120
is the fastest way.

Answer 3

Since no three points on a circle can be collinear, any 3 points will do. How many ways are there to choose 3 points out of 10? Combinations!

C(n, k) = n! / [k! (n-k)!] is the number of combinations of k units that one can choose among n units. In your case n=10 and k=3.

Answer 4

For the first point you have 10 choices. For the second, 9. For the 3rd, 8. So, my vote is for the product of those 3 #s, or 720 total.

Answer 5

this problem is a combination without repetition
with n = 10 and k =3
C( n k ) = n! / ( k! ( n-k )! ) = 120

find out more :