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by Amrabdelnaby » Wed Dec 30, 2015 9:52 am
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices among the 9 points?

20

30

40

70

90
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by GMATGuruNY » Wed Dec 30, 2015 10:01 am
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, How many triangles can be formed whose vertices are among the 9 points?


A. 20
B. 30
C. 40
D. 70
E. 90
One approach:

To form a triangle, we must select a COMBINATION of 3 points from the 9 options contained in the two lines.

Good combinations = total possible combinations - bad combinations.

Total possible combinations:
Number of ways to choose a combination of 3 points from 9 options = 9C3 = (9*8*7)/(3*2*1) = 84.

Bad combinations:
A BAD combination consists of 3 points selected from the SAME LINE (since these 3 points are collinear and thus cannot be used to form a triangle).
Number of ways to choose a combination of 3 points from the 4 options on line one = 4C3 = (4*3*2)/(3*2*1) = 4.
Number of ways to choose a combination of 3 points from the 5 options on line two = 5C3 = (5*4*3)/(3*2*1) = 10.

Thus:
Good combinations = 84-4-10 = 70.

The correct answer is D.
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by Brent@GMATPrepNow » Wed Dec 30, 2015 10:20 am
If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, How many triangles can be formed whose vertices are among the 9 points?


A. 20
B. 30
C. 40
D. 70
E. 90
Another approach:

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, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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 = [spoiler]70 = D[/spoiler]

Cheers,
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by hoppycat » Wed May 31, 2017 5:05 am
Is this a representative question on the GMAT? I was thinking the side of a triangle can't so through more than 2 points on a line. Can it?
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by DavidG@VeritasPrep » Wed May 31, 2017 6:53 am
hoppycat wrote:Is this a representative question on the GMAT? I was thinking the side of a triangle can't so through more than 2 points on a line. Can it?
Yep, you can see questions like this. Imagine the two parallel lines. The triangle is going to have two points on one of the lines. Between the two points there could be other points, but those points obviously won't represent places where a vertex might be drawn.
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by [email protected] » Wed May 31, 2017 8:54 am
Hi hoppycat,

The concepts that are tested in this question will almost certainly show up when you take the Official GMAT. Remember that the GMAT is a 'critical thinking' test - and as you score at higher and higher levels, the Exam will find ways to test you about subjects that you probably know, but in ways that you're probably not used to thinking about. This question could have been phrased in a variety of different ways that didn't involve geometry at all - the geometry just makes you think about the math in a slightly different way.

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by hoppycat » Wed May 31, 2017 12:04 pm
DavidG@VeritasPrep wrote:
hoppycat wrote:Is this a representative question on the GMAT? I was thinking the side of a triangle can't so through more than 2 points on a line. Can it?
Yep, you can see questions like this. Imagine the two parallel lines. The triangle is going to have two points on one of the lines. Between the two points there could be other points, but those points obviously won't represent places where a vertex might be drawn.
perfect thanks!
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by Matt@VeritasPrep » Mon Jun 05, 2017 11:05 pm
hoppycat wrote:Is this a representative question on the GMAT? I was thinking the side of a triangle can't so through more than 2 points on a line. Can it?
Not 100% sure of this, but I think this Q (or one very much like it) is in the one of the Exam Packs from mba.com.
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Re: Counting

by Scott@TargetTestPrep » Fri Feb 21, 2020 8:01 am
Amrabdelnaby wrote: ↑
Wed Dec 30, 2015 9:52 am
 If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices among the 9 points?

20

30

40

70

90
Solution:

We have two scenarios:

The first scenario is that the 1 vertex is on the first line and the 2 other vertices are on the second line.
So the number of ways to create such a triangle is:

4C1 x 5C2 = 4 x (5 x 4)/2 = 40

The second scenario is that 2 vertices are on the first line and the 1 vertex is on the second line. So the number of ways to create such a triangle is:

4C2 x 5C1 = (4 x 3)/2 x 5 = 6 x 5 = 30

Thus, the total number of ways to create the triangle is 40 + 30 = 70.

Alternate Solution:

If it had been so that no three of the 4 + 5 = 9 points were collinear, there would have been 9C3 = 9!/(3!*6!) = (9*8*7)/(3*2) = 84 possible triangles.

However, we are given that four points belong to one line and five points belong to another line. Thus, we must exclude the triple of points chosen entirely from the first group and also the triple of points chosen entirely from the second group.

From the group consisting of four points, 4C3 = 4 triples can be chosen. From the group consisting of five points, 5C3 = 5!/(3!*2!) = (5 x 4)/2 = 10 triples can be chosen. Thus, the 4 + 10 = 14 triples among the 84 triples we calculated earlier does not form a triangle. Taking away those 14 triples, we see that 84 - 14 = 70 triangles are possible.

Answer: D

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