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Intern  Joined: 18 Mar 2012
Posts: 46
GPA: 3.7
If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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2
34 00:00

Difficulty:   25% (medium)

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?

A. 20
B. 30
C. 40
D. 70
E. 90
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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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27
21
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.
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Posts: 220
WE 1: IT 1 Yr
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Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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8
4
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.
##### General Discussion
Senior Manager  Joined: 13 Aug 2012
Posts: 418
Concentration: Marketing, Finance
GPA: 3.23
Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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3
2
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

Combine possiblities: 30 + 40 = 70
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Intern  Joined: 10 Apr 2012
Posts: 22
Concentration: Finance, Economics
GMAT 1: 760 Q50 V44 Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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2
1
See Image below for what my scratch paper looked like on this one
Attachments Gmat_56_resized.jpg [ 101.22 KiB | Viewed 18343 times ]

Manager  Joined: 22 Feb 2009
Posts: 161
Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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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 _________________
.........................................................................
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Intern  Joined: 29 Sep 2013
Posts: 6
Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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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  [#permalink]

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1
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?

Combination problems are located at search.php?search_id=tag&tag_id=52
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Re: If 4 points are indicated on a line and 5 points are indicated on  [#permalink]

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Top Contributor
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.
Attachments 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  [#permalink]

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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.

Answer: D
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_________________ Re: If 4 points are indicated on a line and 5 points are indicated on   [#permalink] 17 Apr 2019, 12:45
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