In a triangle ABC, there are 5, 6 and 4 points, different from the ver

To try to go through each case would be difficult.

First, you could have the case in which you pick one point from each of the 3 sides of the triangle.

Second, you could pick 2 points from one side and a 3rd from the remaining two sides

However, you would have to account for over counting as well, since the corner vertices will be counted as part of two sides.

On the other hand, it is much easier to just subtract out the Unfavorable Combinations.

Since we need 3 vertices/points to make a triangle, first we can find every unique way to make a combination of 3 points out of 18 available points:

(total number of unique ways to make combinations of 3 points out of a total of 18 available points)

—

(Unfavorable combinations in which the 3 points are collinear and can NOT represent a triangle)

Which is equal to:


(18 c 3)

— (no. of ways to have a group of 3 linear points from side AB: 7 c 3)

— (no. of ways to have a group of 3 linear points from side BC: 8 c 3)

— (no. of ways to have a group of 3 linear points from side AC: 6 c 3)

When you calculate the factorials, you get:

816 — (35 + 56 + 20) =

705

*B*

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There are a total of 18 points from which we need three non-colinear ones.

18*17*16/3*2*1 = 3*17*16 = 51*16 = 800+16 = 816

Now we need to eliminate some. C, D, and E are out.

We are only eliminating the ones with three points all from the same side. That's not the majority. A is out.

Answer choice B.