Source: Veritas Prep
Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?
A. 56
B. 70
C. 74
D. 79
E. 84
The OA is C.
Set S contains nine distinct points in the coordinate plane.
This topic has expert replies
-
- Moderator
- Posts: 2205
- Joined: Sun Oct 15, 2017 1:50 pm
- Followed by:6 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
To form a triangle, we must select 3 points such that at most 2 are collinear.BTGmoderatorLU wrote:Source: Veritas Prep
Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?
A. 56
B. 70
C. 74
D. 79
E. 84
Case 1: Select 3 points not on the x-axis
From the 4 points not on the x-axis, the number of ways to choose 3 = 4C3 = (4*3*2)/(3*2*1) = 4.
Case 2: Select 1 point on the x-axis and two points not on the x-axis
From the 5 points on the x-axis, the number of ways to choose 1 = 5C1 = 5.
From the 4 points not on the x-axis, the number of ways to choose 2 = 4C2 = (4*3)/(2*1) = 6.
To combine these options, we multiply:
5*6 = 30.
Case 3: Select 2 points on the x-axis and 1 point not on the x-axis
From the 5 points on the x-axis, the number of ways to choose 2 = 5C2 = (5*4)/(2*1) = 10.
From the 4 points not on the x-axis, the number of ways to choose 1 = 4C1 = 4.
To combine these options, we multiply:
10*4 = 40.
Total ways = Case 1 + Case 2 + Case 3 = 4 + 30 + 40 = 74.
The correct answer is C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
\[? = C\left( {9,3} \right) - C\left( {5,3} \right) = \frac{{9 \cdot 8 \cdot 7}}{{3 \cdot 2}} - \frac{{5 \cdot 4}}{2} = 12 \cdot 7 - 10 = 74\]BTGmoderatorLU wrote:Source: Veritas Prep
Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?
A. 56
B. 70
C. 74
D. 79
E. 84
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7223
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We have 5 (collinear) points that are on the x-axis and 4 points that are not on the x-axis. Since no set of three points in S is collinear except those on the x-axis, we can have the following 3 cases forming a triangle: 1) two points on the x-axis and one point not on the x-axis, 2) one point on the x-axis and two points not on the x-axis, and 3) three points not on the x-axis.BTGmoderatorLU wrote:Source: Veritas Prep
Set S contains nine distinct points in the coordinate plane. If exactly five of the points lie on the x-axis, and if no other set of three points in S is collinear, how many triangles can be formed by taking points in S as vertices?
A. 56
B. 70
C. 74
D. 79
E. 84
Case 1: Two points on the x-axis and one point not on the x-axis
There are 5C2 x 4C1 = 10 x 4 = 40 such triangles in this case.
Case 2: one point on the x-axis and two points not on the x-axis
There are 5C1 x 4C2 = 5 x 6 = 30 such triangles in this case.
Case 3: Three points not on the x-axis
There are 4C3 = 4 such triangles in this case.
Therefore, there are 40 + 30 + 4 = 74 triangles that can be formed.
Answer: C
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews