How many triangles can be formed from 12 points out of which 7 of them are always collinear?

How many triangles can be obtained by joining 12 points, four of which are collinear?

Answer : Total number of points on plane = 12 Triangles can be formed from these points = 12C3

= 220

But 4 points are colinear, the number of triangles can be formed from these points

= 4C3

= 4

We need to subtract 4 from 220 because in the formation of triangles from 4 colinear points are added there.

So no of triangle formed is = 220 – 4

= 216

There are 12 distinct non-collinear points in a same plane, they are points A,B,....L.

How many different triangle can be formed, with criteria one of its vertice must be contain point A?

My attempt:

Because the arrangements didn't need an order, so we can use combination to solve this problem.

Since there are 12 points and we need only to take three of them, so the possibility is

C(12,3) = 220

If there is no criteria, I think this is the total number to create the triangle from the 12 distinct points.

But, how about the numbers of solution if the criteria is required one of its vertice must be point A? Is the total possibility remains the same?

Thanks

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The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is 185.

Explanation:

Total number of triangles formed from 12 points taking 3 at a time = 12C3

But given that out of 12 points, 7 are collinear

So, these seven points will form no triangle.

∴ The required number of triangles = 12C3 – 7C3

= `(12!)/(3!  9!) - (7!)/(3!4!)`

= `(12 xx 11 xx 10 xx 9!)/(3 xx 2 xx 1 xx 9!) - (7 xx 6 xx 5 xx 4!)/(3 xx 2 xx 1 xx 4!)`

= `(12 xx 11 xx 10)/(3 xx 2) - (7 xx 6 xx 5)/(3 xx 2)`

= 220 – 35

= 185