points in a plane

[sanju09]

There are 20 points in a plane, and 5 points of them are collinear.
Q 1
How many triangles can be made using these points as vertices?
A. 20C3
B. 20C3 - 5C3
C. 15C3
D. 15C3 + 5C3
E. 15C3 * 5C3

Q 2
How many unique straight lines can be drawn passing through at least 2 of these 20 points?
A. 20C2
B. 20C2 - 5C2
C. 20C2 - 5C2 + 1
D. 15C2 * 5C2
E. 15C2 * 5C2 + 1

[Brent@GMATPrepNow]

There are 20 points in a plane, and 5 points of them are collinear.
Q 1
How many triangles can be made using these points as vertices?
A. 20C3
B. 20C3 - 5C3
C. 15C3
D. 15C3 + 5C3
E. 15C3 * 5C3

If there were no collinear points, then any 3 points chosen from the 20 points would define a unique triangle. There are 20C3 ways to accomplish this.
However, if we choose 3 collinear points (3 points on a line) then this does not form a triangle and, as such, these types of "triangles" should not be included in our calculation.
In how may ways can we choose 3 points that are collinear? There are 5 collinear points and we must choose 3 of them. We can accomplich this in 5C3 ways.

So, the total number of triangles is 20C3 - 5C3 (B)

How many straight lines can be drawn passing through at least 2 of these 20 points?
A. 20C2
B. 20C2 - 5C2
C. 20C2 - 5C2 + 1
D. 15C2 * 5C2
E. 15C2 * 5C2 + 1

Note: I will assume that the question is asking for the number of unique lines.
Any two points will define a line.
If there were no collinear points, then any 2 points chosen from the 20 points would define a unique line. There are 20C2 ways to accomplish this.
However, any two points chosen from the 5 collinear points will yield the same line. So, lines of this nature should be subtracted from our original calculation.
We can choose 2 points from the 5 collinear points in 5C2 ways.
At this point, it appears that our answer is 20C2 - 5C2. However, we should note that 5C2 represents all of the ways in which we can select 2 points that all define the same line. By subtracting all 5C2 instances, we are excluding one of the possible lines. So, we must add 1 to the calculation.
The total number of ways is 20C2 - 5C2 + 1 (C)

[sanju09]

Thank you Brent for your wonderful elucidations, and one for you pointed out the missing adjective in Q 2, 'unique'; it's now there