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If 4 points are indicated on a line and 5 points are indicated on
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16 May 2012, 06:52
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Question Stats:
74% (01:37) correct 26% (01:51) wrong based on 452 sessions
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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?
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16 May 2012, 07:58
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alexpavlos wrote:
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
Tried doing a few options, but couldn't get it right and the answer provided left me more confused!
Approach #1:
There are two types of triangles possible: With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\); With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);
Total: \(30+40=70\).
Answer: D.
Approach #2:
All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear. \(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).
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16 May 2012, 07:04
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say line A with 4 points and line B with 5 points. and triangle is PQR. If P is on line A and QR on line B we can have 4C1*5C2=40 triangles. If P is on line B and QR on line A we can have 4C2*5C1=30 triangles. Total =70 triangles.
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29 Dec 2012, 01:27
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Not that fun this question I wonder how often I would see such a question on test day.
Solution:
* * * * * (line B)
* * * * (line A)
Choose 2 points from line A and a point from line B: = 4!/2!2! * 5!/1!4! = 30 Choose 1 point from line A and 2 points from line B: = 4!/1!3! * 5!/2!3! = 4 * 10 = 40
Re: If 4 points are indicated on a line and 5 points are indicated on
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19 Jul 2014, 22:47
Bunuel wrote:
alexpavlos wrote:
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
Tried doing a few options, but couldn't get it right and the answer provided left me more confused!
Approach #1: There are two types of triangles possible: With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\); With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);
Total: \(30+40=70\).
Answer: D.
Approach #2:
All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear. \(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).
Answer: D.
Hope it's clear.
I really like your 2nd approach _________________
......................................................................... +1 Kudos please, if you like my post
Re: If 4 points are indicated on a line and 5 points are indicated on
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06 Aug 2015, 14:54
Bunuel wrote:
alexpavlos wrote:
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
Tried doing a few options, but couldn't get it right and the answer provided left me more confused!
Approach #1: There are two types of triangles possible: With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\); With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);
Total: \(30+40=70\).
Answer: D.
Approach #2:
All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear. \(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).
Answer: D.
Hope it's clear.
Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?
Re: If 4 points are indicated on a line and 5 points are indicated on
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06 Aug 2015, 15:29
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dukenukem wrote:
Bunuel wrote:
alexpavlos wrote:
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
Tried doing a few options, but couldn't get it right and the answer provided left me more confused!
Approach #1: There are two types of triangles possible: With two vertices on the line with 4 points and the third vertex on the line with 5 points --> \(C^2_4*C^1_5=30\); With two vertices on the line with 5 points and the third vertex on the line with 4 points --> \(C^2_5*C^1_4=40\);
Total: \(30+40=70\).
Answer: D.
Approach #2:
All different 3 points out of total 4+5=9 points will create a triangle EXCEPT those 3 points which are collinear. \(C^3_{9}-(C^3_4+C^3_5)=84-(4+10)=70\) (where \(C^3_4\) and \(C^3_4\) are # of different 3 collinear points possible from the line with 4 points and the line with 5 points, respectively).
Answer: D.
Hope it's clear.
Thanks Bunuel. That was great. Do you by any chance have other problems similar to this that you know of?
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Re: If 4 points are indicated on a line and 5 points are indicated on
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06 Feb 2017, 09:47
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Attached is a visual that should help. Bunuel's method is the most elegant, but here is an alternative strategy if the other solution is not clear.
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Screen Shot 2017-02-06 at 8.46.05 AM.png [ 117.98 KiB | Viewed 11179 times ]
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Re: If 4 points are indicated on a line and 5 points are indicated on
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14 Mar 2018, 16:09
alex1233 wrote:
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
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.
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17 Apr 2019, 12:45
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74% (01:37) correct 
I wonder how often I would see such a question on test day.
