how many triangles can be formed

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Needgmat: Master | Next Rank: 500 Posts; Posts: 234; Joined: Tue May 31, 2016 1:40 am; Thanked: 3 times

by Needgmat » Tue Aug 30, 2016 8:12 am

If 4 points 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

OAD

Please explain.

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Source: Beat The GMAT — Problem Solving | Tue Aug 30, 2016 8:12 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue Aug 30, 2016 9:58 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

When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer

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 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,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Tue Aug 30, 2016 10:03 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

Alternate 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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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Sep 01, 2016 5:32 pm

Needgmat wrote:If 4 points 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

OAD

Please explain.

I'd think of it this way: we need ONE point on the top line (whichever line that is) and TWO points on the bottom line (whichever line that is).

If line A is the top line, we have 4 choices of point from that line and (5 choose 2), or 10, choices from line B, for a total of 4 * 10 => 40 choices.

If line B is the top line, we have 5 choices of point and (4 choose 2), or 6, choices from line A, for a total of 5 * 6 => 30 choices.

40 + 30 => 70, so that's our answer!

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8137; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Wed Mar 14, 2018 3:26 pm

Needgmat wrote:If 4 points 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

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.

Answer: D

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