Triangles in 20 non-straight-line "dots"

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August 22nd 2010, 06:07 AM
grottvald

Triangles in 20 non-straight-line "dots"

Twenty points/dots are given so that the three of them are never in a straight line. How many triangles can be formed with corners(as in vertex i think?) in the dots?
August 22nd 2010, 06:21 AM
Plato

Calculate the number of ways to choose three points from twenty.
August 22nd 2010, 09:36 AM
grottvald

C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)
August 22nd 2010, 09:57 AM
Plato

Quote:

Originally Posted by grottvald

C(20,3) ? But that results in a very big number that seems unlikely to be correct

It is not a large number at all: $\displaystyle \binom{20}{3}=\frac{20!}{3!\cdot 17!}=1140$
August 22nd 2010, 10:58 AM
Archie Meade

Quote:

Originally Posted by grottvald

C(20,3) ? But that results in a very big number that seems unlikely to be correct (Worried)

You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.

The number of ways to do this is $\binom{19}{2}=171$

Therefore, there are 171 triangles that can be drawn which include the leftmost point.

If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.

These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.

Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.

hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.
August 22nd 2010, 12:26 PM
grottvald

Quote:

Originally Posted by Archie Meade

You may be thinking of non-overlapping triangles in the picture with 20 dots.
That's a different situation.
Imagine the dots are placed apart from left to right, not on a straight line.
Pick the leftmost point.
To make a triangle, you can pick any 2 of the remaining 19 dots.

The number of ways to do this is $\binom{19}{2}=171$

Therefore, there are 171 triangles that can be drawn which include the leftmost point.

If you move on to the next point to the right and exclude the point previously chosen,
then you can draw another $\binom{18}{2}=153$ triangles.

These triangles do not include any of the previous 171,
since these 153 omit the leftmost point.

Notice that these triangles typically share sides of other triangles,
but they are made of 3 distinct points, hence the triangles are being counted only once.

hence, there are $\binom{19}{2}+\binom{18}{2}+\binom{17}{2}+........ .+\binom{2}{2}=\binom{20}{3}$ triangles.

Thank you so much! You explained it in an very easy and beautiful way. Thanks again!