vinay1983 wrote:There are 12 points on a plane, 6 of which are collinear. How many distinct triangles can be formed using these 12 points as vertices?
A. 220
B. 200
C. 185
D. 35
E. 20
If it were the case that no three lines were collinear, then a selection of
any 3 points would create a distinct triangle.
If the 3 selected point are collinear, then no triangle is formed.
So, for this question, if we select any 3 points, there's a chance that a triangle is not formed. So, take all of the 3-point selections, and subtract those selections that don't yield a triangle.
In other words, the TOTAL number of distinct triangles = (
number of ways to select 3 points from all 12) - (
number of ways to select 3 points from the 6 collinear points)
Number of ways to select 3 points from all 12
Since the order in which we select the 3 points does not matter, we can use combinations.
We can select 3 points from 12 in 12C3 ways = 220 ways
Aside: if anyone is interested, we have a free video on calculating combinations (like 12C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Number of ways to select 3 points from the 6 collinear points
We can select 3 points from 6 in 6C3 ways = 20 ways
So, the TOTAL number of distinct triangles =
220 -
20 = [spoiler]200 = B[/spoiler]
Cheers,
Brent
