In a plane, there are two parallel lines. One line has 5 points and an

There are 2 parallel lines. 5 points on one line, 4 points on the other line. To make a triangle, you need 2 points from one line and 1 point from the other line.

# of triangles = (# of ways to select 2 out of 5 points) and (# of ways to select 1 out of 4 points) or (# of ways to select 1 out of 5 points) and (# of ways to select 2 out of 4 points)

# of ways to select 2 out of 5 points: 5C2=10
# of ways to select 1 out of 4 points: 4C1=4
# of ways to select 1 out of 5 points: 5C1=5
# of ways to select 2 out of 4 points: 4C2=6

# of triangles = 10*4+5*6 = 70

Answer: B

Hey Everyone,

Official solution to the question has been posted.

Regards,
Ashutosh

There are two parallel lines.

One line has five points and another has 4 points in it.

So while selecting points we can't select all the three points from the same line, since it will be colinear
Hence we will select two points from one line and other from second line, which gives us:
5C2*4C1 + 4C2*5C1 = 70

There are two ways in which we can create a triangle.
#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.

#1) Select 2 points from the 5-point line and select 1 point from the 4-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 5-point line
Since the order of the 2 selected points does not matter, we can use combinations.
We can select 2 points from 5 points in 5C2 = 10 ways.

If anyone is interested, a video on calculating combinations (like 5C2) in your head can be found at the bottom of this post

Stage 2: Select 1 point from the 4-point line.
We can complete this stage in 4 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (10)(4) ways (= 40 ways)

#2) Select 2 points from the 4-point line and select 1 point from the 5-point line.
Take this task and break it into stages.

Stage 1: Select 2 points from the 4-point line
We can select 2 points from 4 points in 4C2 = 6 ways.

Stage 2: Select 1 point from the 5-point line.
We can complete this stage in 5 ways

By the Fundamental Counting Principle (FCP) we can complete the 2 stages in (6)(5) ways (= 30 ways)
-------------------------------------------------------------

So, the TOTAL number of triangles = 40 + 30
= 70

Answer: B

Cheers,
Brent

RELATED VIDEO

Solution:

We can let line A be the line with 5 points and line B the line with 4 points. So we can form a triangle by choosing 1 point on line A and 2 points on line B, or by choosing 2 points on line A and 1 point on line B. In the former case, we have 5C1 x 4C2 = 5 x 6 = 30 ways of forming such a triangle, and in the latter case, we have 5C2 x 4C1 = 10 x 4 = 40 ways of forming such a triangle. Therefore, we have a total of 30 + 40 = 70 ways to form a triangle.

Alternate solution:

The number of ways to choose 3 points from 9 is 9C3 = 9! / (3!6!) = (9 x 8 x 7) / (3 x 2) = 3 x 4 x 7 = 84. However, we can’t have all 3 points on the same line (since the 3 points can’t be collinear). The number of ways of choosing 3 points from the line with 5 points is 5C3 = 5! / (3!2!) = (5 x 4 x 3) / (3 x 2) = 5 x 2 = 10 and similarly, the number of ways of choosing all 3 points from the line with 4 points is 4C3 = 4. Therefore, we have a total of 84 - 10 - 4 = 70 ways forming a triangle.

Answer: B

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.